Solutions Manual, Linear Algebra Theory And Applications F.12 Exercises
1
Exercises
Solutions Manual, Linear Algebra Theory And
Applications
F.12
Exercises
1.6
1. Let z = 5 + i9. Find z −1 .
−1
(5 + i9)
=
5
106
−
9
106 i
2. Let z = 2 + i7 and let w = 3 − i8. Find zw, z + w, z 2 , and w/z.
62 + 5i, 5 − i, −45 + 28i, and − 50
53 −
37
53 i.
3. Give the complete solution to x4 + 16 = 0.
√
√
√
√
x4 + 16 = 0, Solution is: (1 − i) 2, − (1 + i) 2, − (1 − i) 2, (1 + i) 2.
4. Graph the complex cube roots of 8 in the complex plane. Do the same for the four
fourth roots of 16.
√
√
The cube roots are the solutions to z 3 + 8 = 0, Solution is: i 3 + 1, 1 − i 3, −2
The fourth roots are the solutions to z 4 + 16 = 0, Solution is:
√
√
√
(1 − i) 2, − (1 + i) 2, − (1 − i) 2, (1 + i)
√
2. When you graph these, you will have three equally spaced points on the circle of
radius 2 for the cube roots and you will have four equally spaced points on the circle
of radius 2 for the fourth roots. Here are pictures which should result.
5. If z is a complex number, show there exists ω a complex number with |ω| = 1 and
ωz = |z| .
z
If z = 0, let ω = 1. If z 6= 0, let ω =
|z|
n
6. De Moivre’s theorem says [r (cos t + i sin t)] = rn (cos nt + i sin nt) for n a positive
integer. Does this formula continue to hold for all integers, n, even negative integers?
Explain.
Yes, it holds for all integers. First of all, it clearly holds if n = 0. Suppose now that
n is a negative integer. Then −n > 0 and so
n
[r (cos t + i sin t)] =
1
1
−n = −n
r
(cos
(−nt)
+ i sin (−nt))
[r (cos t + i sin t)]
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
2
Exercises
=
=
rn
rn (cos (nt) + i sin (nt))
=
(cos (nt) − i sin (nt))
(cos (nt) − i sin (nt)) (cos (nt) + i sin (nt))
n
r (cos (nt) + i sin (nt))
because (cos (nt) − i sin (nt)) (cos (nt) + i sin (nt)) = 1.
7. You already know formulas for cos (x + y) and sin (x + y) and these were used to prove
De Moivre’s theorem. Now using De Moivre’s theorem, derive a formula for sin (5x)
and one for cos (5x).
sin (5x) = 5 cos4 x sin x − 10 cos2 x sin3 x + sin5 x
cos (5x) = cos5 x − 10 cos3 x sin2 x + 5 cos x sin4 x
8. If z and w are two complex numbers and the polar form of z involves the angle θ while
the polar form of w involves the angle φ, show that in the polar form for zw the angle
involved is θ + φ. Also, show that in the polar form of a complex number, z, r = |z| .
You have z = |z| (cos θ + i sin θ) and w = |w| (cos φ + i sin φ) . Then when you multiply
these, you get
=
=
|z| |w| (cos θ + i sin θ) (cos φ + i sin φ)
|z| |w| (cos θ cos φ − sin θ sin φ + i (cos θ sin φ + cos φ sin θ))
|z| |w| (cos (θ + φ) + i sin (θ + φ))
9. Factor x3 + 8 as a product of linear factors.
√
√
x3 + 8 = 0, Solution is: i 3 + 1, 1 − i 3, −2 and so this polynomial equals
√
√ (x + 2) x − i 3 + 1
x− 1−i 3
10. Write x3 + 27 in the form (x + 3) x2 + ax + b where x2 + ax + b cannot be factored
any more using only real numbers.
x3 + 27 = (x + 3) x2 − 3x + 9
11. Completely factor x4 + 16 as a product of linear factors.
√
√
√
√
x4 + 16 = 0, Solution is: (1 − i) 2, − (1 + i) 2, − (1 − i) 2, (1 + i) 2. These are
just the fourth roots of −16. Then to factor, this you get
√ √ x − (1 − i) 2
x − − (1 + i) 2 ·
√ √ x − − (1 − i) 2
x − (1 + i) 2
12. Factor x4 + 16 as the product of two quadratic polynomials each of which cannot be
factored further without using complex numbers.
√
√
x4 + 16 = x2 − 2 2x + 4 x2 + 2 2x + 4 . You can use the information in the
preceding problem. Note that (x − z) (x − z) has real coefficients.
13. If z, w are complex numbersP
prove zw =Pzw and then show by induction that z1 · · · zm =
m
m
z1 · · · zm . Also verify that k=1 zk = k=1 zk . In words this says the conjugate of a
product equals the product of the conjugates and the conjugate of a sum equals the
sum of the conjugates.
(a + ib) (c + id) = ac − bd + i (ad + bc) = (ac − bd) − i (ad + bc)
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
3
Exercises
(a − ib) (c − id) = ac − bd − i (ad + bc) which is the same thing. Thus it holds for
a product of two complex numbers. Now suppose you have that it is true for the
product of n complex numbers. Then
z1 · · · zn+1 = z1 · · · zn zn+1
and now, by induction this equals
z1 · · · zn zn+1
As to sums, this is even easier.
n
X
(xj + iyj ) =
j=1
=
n
X
j=1
xj − i
n
X
xj + i
j=1
n
X
yj =
j=1
n
X
j=1
xj − iyj =
n
X
yj
j=1
n
X
(xj + iyj ).
j=1
14. Suppose p (x) = an xn + an−1 xn−1 + · · · + a1 x + a0 where all the ak are real numbers.
Suppose also that p (z) = 0 for some z ∈ C. Show it follows that p (z) = 0 also.
You just use the above problem. If p (z) = 0, then you have
p (z) = 0 = an z n + an−1 z n−1 + · · · + a1 z + a0
= an z n + an−1 z n−1 + · · · + a1 z + a0
= an z n + an−1 z n−1 + · · · + a1 z + a0
= an z n + an−1 z n−1 + · · · + a1 z + a0
= p (z)
15. I claim that 1 = −1. Here is why.
−1 = i2 =
q
√
√ √
2
−1 −1 = (−1) = 1 = 1.
This is clearly a remarkable result but is there something wrong with it? If so, what
is wrong?
√
Something is wrong. There is no single −1.
16. De Moivre’s theorem is really a grand thing. I plan to use it now for rational exponents,
not just integers.
1 = 1(1/4) = (cos 2π + i sin 2π)
1/4
= cos (π/2) + i sin (π/2) = i.
Therefore, squaring both sides it follows 1 = −1 as in the previous problem. What
does this tell you about De Moivre’s theorem? Is there a profound difference between
raising numbers to integer powers and raising numbers to non integer powers?
It doesn’t work. This is because there are four fourth roots of 1.
17. Show that C cannot be considered an ordered field. Hint: Consider i2 = −1.
It is clear that 1 > 0 because 1 = 12 . (In general a2 > 0. This is clear if a > 0. If a < 0,
then adding −a to both sides, yields that 0 < −a. Also recall that −a = (−1) a and
2
2
2
that (−1) = 1. Therefore, (−a) = (−1) a2 = a2 > 0. ) Now it follows that if C can
be ordered, then −1 > 0, but this is a problem because it implies that 0 > 1 = 12 > 0.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
4
Exercises
18. Say a + ib < x + iy if a < x or if a = x, then b < y. This is called the lexicographic
order. Show that any two different complex numbers can be compared with this order.
What goes wrong in terms of the other requirements for an ordered field.
From the definition of this funny order, 0 < i and so if this were an order, you would
need to have 0 < i2 = −1. Now add 1 to both sides and obtain 0 > 1 = 12 > 0, a
contradiction.
19. With the order of Problem 18, consider for n ∈ N the complex number 1 − n1 . Show
that with the lexicographic order just described, each of 1 − im is an upper bound to
all these numbers. Therefore, this is a set which is “bounded above” but has no least
upper bound with respect to the lexicographic order on C.
This follows from the definition. 1 − im > 1 − 1/n for each m. Therefore, if you
consider the numbers 1 − n1 you have a nonempty set which has an upper bound but
no least upper bound.
F.13
Exercises
1.11
1. Give the complete solution to the system of equations, 3x − y + 4z = 6, y + 8z = 0,
and −2x + y = −4.
x = 2 − 4t, y = −8t, z = t.
2. Give the complete solution to the system of equations, x+ 3y + 3z = 3, 3x+ 2y + z = 9,
and −4x + z = −9.
x = y = 2, z = −1
3. Consider the system −5x + 2y − z = 0 and −5x − 2y − z = 0. Both equations equal
zero and so −5x + 2y − z = −5x − 2y − z which is equivalent to y = 0. Thus x and
z can equal anything. But when x = 1, z = −4, and y = 0 are plugged in to the
equations, it doesn’t work. Why?
These are invalid row operations.
4. Give the complete solution to the system of equations, x+2y +6z = 5, 3x+2y +6z = 7
,−4x + 5y + 15z = −7.
No solution.
5. Give the complete solution to the system of equations
x + 2y + 3z
−4x + 5y + z
=
=
5, 3x + 2y + z = 7,
−7, x + 3z = 5.
x = 2, y = 0, z = 1.
6. Give the complete solution of the system of equations,
x + 2y + 3z
−4x + 5y + 5z
= 5, 3x + 2y + 2z = 7
= −7, x = 5
No solution.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
5
Exercises
7. Give the complete solution of the system of equations
x + y + 3z
=
−4x + 9y + z
=
2, 3x − y + 5z = 6
−8, x + 5y + 7z = 2
x = 2 − 2t, y = −t, z = t.
8. Determine a such that there are infinitely many solutions and then find them. Next
determine a such that there are no solutions. Finally determine which values of a
correspond to a unique solution. The system of equations is
3za2 − 3a + x + y + 1 = 0
3x − a − y + z a2 + 4 − 5 = 0
za2 − a − 4x + 9y + 9 = 0
If a = 1, there are infinitely many solutions of the form x = 2 − 2t, y = −t, z = t. If
a = −1, t then there are no solutions. If a is anything else, there is a unique solution.
2a
1
1
x=
,y = −
,z =
a+1
a+1
a+1
9. Find the solutions to the following system of equations for x, y, z, w.
y + z = 2, z + w = 0, y − 4z − 5w = 2, 2y + z − w = 4
x = t, y = s + 2, z = −s, w = s
10. Find all solutions to the following equations.
x+y+z
=
2, z + w = 0,
2x + 2y + z − w
=
4, x + y − 4z − 5z = 2
x = −t + s + 2, y = t, z = −s, w = s, where s, t are each in F.
F.14
Exercises
1.14
1. Verify all the properties 1.11-1.18.
2. Compute 5 (1, 2 + 3i, 3, −2) + 6 (2 − i, 1, −2, 7) .
3. Draw a picture of the points in R2 which are determined by the following ordered
pairs.
(a)
(b)
(c)
(d)
(1, 2)
(−2, −2)
(−2, 3)
(2, −5)
4. Does it make sense to write (1, 2) + (2, 3, 1)? Explain.
5. Draw a picture of the points in R3 which are determined by the following ordered
triples. If you have trouble drawing this, describe it in words.
(a) (1, 2, 0)
(b) (−2, −2, 1)
(c) (−2, 3, −2)
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
6
Exercises
F.15
Exercises
1.17
1. Show that (a · b) =
1
4
h
i
2
2
|a + b| − |a − b| .
2
2. Prove from the axioms of the inner product the parallelogram identity, |a + b| +
|a − b|2 = 2 |a|2 + 2 |b|2 .
P
3. For a, b ∈ Rn , define a · b ≡ nk=1 β k ak bk where β k > 0 for each k. Show this satisfies
the axioms of the inner product. What does the Cauchy Schwarz inequality say in
this case.
The product satisfies all axioms for the inner product so the Cauchy Schwarz inequality
holds.
4. In Problem 3 above, suppose you only know β k ≥ 0. Does the Cauchy Schwarz inequality still hold? If so, prove it.
Yes, it does. You don’t need the part which says that the only way a · a = 0 is for
a = 0 in the argument for the Cauchy Schwarz inequality.
5. Let f, g be continuous functions and define
Z
f ·g ≡
1
f (t) g (t)dt
0
show this satisfies the axioms of a inner product if you think of continuous functions
in the place of a vector in Fn . What does the Cauchy Schwarz inequality say in this
case?
The only part which is not obvious for the axioms is the one which says that if
Z
1
0
2
|f | = 0
then f = 0. However, this is obvious from continuity considerations.
6. Show that if f is a real valued continuous function,
Z
b
f (t) dt
a
Z
b
f (t) dt
a
≤
!2
Z
b
≤ (b − a)
2
1 dt
a
1/2
= (b − a)
Z
!1/2
Z
a
b
b
2
f (t) dt.
a
Z
a
b
2
|f (t)| dt
2
|f (t)| dt
!1/2
!1/2
which yields the desired inequality when you square both sides.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
7
Exercises
F.16
Exercises
2.2
1. In 2.1 - 2.8 describe −A and 0.
2. Let A be an n×n matrix. Show A equals the sum of a symmetric and a skew symmetric
matrix.
A=
A+AT
2
+
A−AT
2
3. Show every skew symmetric matrix has all zeros down the main diagonal. The main
diagonal consists of every entry of the matrix which is of the form aii . It runs from
the upper left down to the lower right.
You know that Aij = −Aji . Let j = i to conclude that Aii = −Aii and so Aii = 0.
4. Using only the properties 2.1 - 2.8 show −A is unique.
Suppose that B also works. Then
−A = −A + (A + B) = (−A + A) + B = B
5. Using only the properties 2.1 - 2.8 show 0 is unique.
If 00 is another additive identity, then 00 = 0 + 00 = 0
6. Using only the properties 2.1 - 2.8 show 0A = 0. Here the 0 on the left is the scalar 0
and the 0 on the right is the zero for m × n matrices.
0A = (0 + 0) A = 0A + 0A. Now add the additive inverse of 0A to both sides.
7. Using only the properties 2.1 - 2.8 and previous problems show (−1) A = −A.
0 = 0A = (1 + (−1)) A = A + (−1) A. Hence, (−1) A is the unique additive inverse of
A. Thus −A = (−1) A.
8. Prove 2.17.
T
(AB)ij ≡ (AB)ji =
P
k
Ajk Bki =
P
k
T T
Bik
Akj = B T AT
9. Prove that Im A = A where A is an m × n matrix.
P
(Im A)ij ≡ k Iik Akj = Aij and so Im A = A.
ij
. Hence the formula holds.
n
10. Let A and
y ∈ Rm . Show (Ax, y)Rm =
be a real m × n matrix and let x ∈ R and
k
T
x,A y Rn where (·, ·)Rk denotes the dot product in R .
P
P P
(Ax, y) = i (Ax)i yi = i k Aik xk yi
P
P
P P
x,AT y = k xk i AT ki yi = k i xk Aik yi , the same as above. Hence the two
are equal.
T
11. Use the result of Problem 10 to verify directly that (AB) = B T AT without making
any reference to subscripts.
(AB)T x, y ≡ (x, (AB) y) = AT x,By = B T AT x, y . Since this holds for every
x, y, you have for all y
T
(AB) x − B T AT x, y
T
Let y = (AB) x − B T AT x. Then since x is arbitrary, the result follows.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
8
Exercises
12. Let x = (−1, −1, 1) and y = (0, 1, 2) . Find xT y and xyT if possible.
−1
0 −1 −2
xT y = −1 0 1 2 = 0 −1 −2
1
0 1
2
0
xyT = −1 −1 1 1 = 1
2
13. Give
1
1
1
1
an example
1
1
1
−1
1
−1
1
1
1
14. Let A = −2
1
if possible.
of matrices, A, B, C such that B 6= C, A 6= 0, and yet AB = AC.
−1
0 0
=
1
0 0
1
0 0
=
−1
0 0
1
1
1 −3
1
−1
−2
−1 , B =
0 . Find
, and C = −1 2
2 1 −2
2
−3 −1 0
(a) AB
(b) BA
(c) AC
(d) CA
(e) CB
(f) BC
15. Consider the following digraph.
1
2
3
4
Write the matrix associated with this digraph and find the number of ways to go from
3 to 4 in three steps.
0 1 1 0
1 0 0 1
The matrix for the digraph is
1 1 0 2
0 1 0 1
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
9
Exercises
0
1
1
0
1
0
1
1
1
0
0
0
3
0
1
1
= 3
4
2
1
1
5
2
5
3
2
0
1
1
4
5
8
3
Thus it appears that there are 8 ways to do this.
16. Show that if A−1 exists for an n × n matrix, then it is unique. That is, if BA = I and
AB = I, then B = A−1 .
From the given equations, multiply on the right by A−1 . Then B = A−1 .
17. Show (AB)
−1
= B −1 A−1 .
ABB −1 A−1 = AIA−1 = I
B −1 A−1 AB = B −1 IB = I
Then by the definition of the inverse and its uniqueness, it follows that (AB)
and
(AB)−1 = B −1 A−1
−1
exists
−1
T
18. Show that if A is an invertible n × n matrix, then so is AT and AT
= A−1 .
T
T
AT A−1 = A−1 A = I
T
T
A−1 AT = AA−1 = I Then from the definition of the inverse and its uniqueness,
T
−1
it follows that AT
exists and equals A−1 .
19. Show that if A is an n × n invertible matrix and x is a n × 1 matrix such that Ax = b
for b an n × 1 matrix, then x = A−1 b.
Multiply both sides on the left by A−1 .
20. Give an example of a matrix A such that A2 = I and yet A 6= I and A 6= −I.
2 0 1
1 0
=
1 0
0 1
21. Give an example of matrices, A, B such that neither A nor B equals zero and yet
AB = 0.
1 1
1 −1
0 0
=
1 1
−1 1
0 0
x1 − x2 + 2x3
x1
2x3 + x1
in the form A x2 where A is an appropriate matrix.
22. Write
x3
3x3
3x4 + 3x2 + x1
x4
1 −1 2 0
x1
x1 − x2 + 2x3
1 0 2 0 x2
x1 + 2x3
0 0 3 0 x3 =
3x3
1 3 0 3
x4
x1 + 3x2 + 3x4
23. Give another example other than the one given in this section of two square matrices,
A and B such that AB 6= BA.
Almost anything works.
1 2
1 2
5 2
=
3 4
2 0
11 6
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
10
Exercises
1
2
2
0
1
3
2
4
=
7
2
10
4
24. Suppose A and B are square matrices of the same size. Which of the following are
correct?
2
(a) (A − B) = A2 − 2AB + B 2
Note this.
2
(b) (AB) = A2 B 2 Not this.
(c) (A + B)2 = A2 + 2AB + B 2 Not this.
2
(d) (A + B) = A2 + AB + BA + B 2 This is all right.
(e) A2 B 2 = A (AB) B This is all right.
3
(f) (A + B) = A3 + 3A2 B + 3AB 2 + B 3 Not this.
(g) (A + B) (A − B) = A2 − B 2 Not this.
(h) None of the above. They are all wrong.
(i) All of the above. They are all right.
−1 −1
25. Let A =
. Find all 2 × 2 matrices, B such that AB = 0.
3
3
−1 −1
x y
−x − z −w − y
=
3
3
z w
3x + 3z 3w + 3y
−z −w
, z, w arbitrary.
z
w
26. Prove that if A−1 exists and Ax = 0 then x = 0.
Multiply on the left by A−1 .
27. Let
Find
1
2
1
28. Let
Find
1
2
1
29. Let
1 2
A= 2 1
1 0
3
4 .
2
3
4 .
2
A−1 if possible. If A−1 does not exist, determine why.
−1
2 3
−2 4 −5
1 4 = 0
1 −2
0 2
1 −2 3
1 0
A= 2 3
1 0
A−1 if possible. If A−1 does not exist, determine why.
−1
0 3
−2 0
3
3 4 = 0 13 − 32
0 2
1 0 −1
1 2
A= 2 1
4 5
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
3
4 .
10
7KH6D\ORU)RXQGDWLRQ
11
Exercises
Find
1
2
4
30. Let
Find
1
1
2
1
F.17
A−1 if possible. If A−1 does not
2 3
1 4 , row echelon form:
5 10
exist, determine why.
1 0 35
0 1 23 A has no inverse.
0 0 0
1
1
A=
2
1
2 0
1 2
1 −3
2 1
2
0
2
2
A−1 if possible. If A−1 does not exist, determine why.
−1
1
1
−1 12
2 0 2
2
2
1
− 12 − 25
1 2 0
2
= 3
−1 0
1 −3 2
0
1
1
9
2 1 2
−2 − 43
4
4
Exercises
2.7
1. Show the map T : Rn → Rm defined by T (x) = Ax where A is an m × n matrix and
x is an m × 1 column vector is a linear transformation.
This follows from matrix multiplication rules.
2. Find the matrix for the linear transformation which rotates every vector in R2 through
an angle of π/3.
√ 1
cos (π/3) − sin (π/3)
− 21 3
2
√
=
1
1
sin (π/3) cos (π/3)
2 3
2
3. Find the matrix for the linear transformation which rotates every vector in R2 through
an angle of π/4.
√ 1√
cos (π/4) − sin (π/4)
2 − 21√ 2
2
√
=
1
1
sin (π/4) cos (π/4)
2 2
2 2
4. Find the matrix for the linear transformation which rotates every vector in R2 through
an angle of −π/3.
√ 1
1
cos (−π/3) − sin (−π/3)
2√
2 3
=
1
sin (−π/3) cos (−π/3)
− 12 3
2
5. Find the matrix for the linear transformation which rotates every vector in R2 through
an angle of 2π/3.
√ 2 cos (π/3) −2 sin (π/3)
1 − 3
√
=
2 sin (π/3) 2 cos (π/3)
3
1
6. Find the matrix for the linear transformation which rotates every vector in R2 through
an angle of π/12. Hint: Note that π/12 = π/3 − π/4.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
12
Exercises
cos (π/3) − sin (π/3)
cos (−π/4) − sin (−π/4)
sin (π/3) cos (π/3)
sin (−π/4) cos (−π/4)
√
√
√ √ 1√ √
1
1
1
− 4 2√3
4 √2√3 + 4 √2
4 √2√
=
1
1
1
1
4 2 3− 4 2
4 2 3+ 4 2
7. Find the matrix for the linear transformation which rotates every vector in R2 through
an angle of 2π/3 and then reflects across the x axis.
√ 1
−√
cos (2π/3) − sin (2π/3)
1 0
− 12 3
2
=
1
sin (2π/3) cos (2π/3)
0 −1
− 21 3
2
8. Find the matrix for the linear transformation which rotates every vector in R2 through
an angle of π/3 and then reflects across the x axis.
√ 1
− 12 3
1 0
cos (π/3) − sin (π/3)
2
√
=
0 −1
sin (π/3) cos (π/3)
− 21 3
− 12
9. Find the matrix for the linear transformation which rotates every vector in R2 through
an angle of π/4 and then reflects across the x axis.
√ 1√
1
1 0
cos (π/4) − sin (π/4)
2
−
2
2
√
√2
=
1
1
0 −1
sin (π/4) cos (π/4)
−2 2 −2 2
10. Find the matrix for the linear transformation which rotates every vector in R2 through
an angle of π/6 and then reflects across the x axis followed by a reflection across the
y axis.
1√
1
−1 0
1 0
cos (π/6) − sin (π/6)
−2 3
2√
=
0 1
0 −1
sin (π/6) cos (π/6)
− 21
− 21 3
11. Find the matrix for the linear transformation which reflects every vector in R2 across
the x axis and then rotates every vector through an angle of π/4.
√ 1√
1
cos (π/4) − sin (π/4)
1 0
2
2
2
√
√2
=
1
1
sin (π/4) cos (π/4)
0 −1
2
−
2
2 2
12. Find the matrix for the linear transformation which rotates every vector in R2 through
an angle of π/4 and next reflects every vector across the x axis. Compare with the
above problem.
√ 1√
1
1 0
cos (π/4) − sin (π/4)
2
−
2
2
√
√2
=
1
1
0 −1
sin (π/4) cos (π/4)
−2 2 −2 2
13. Find the matrix for the linear transformation which reflects every vector in R2 across
the x axis and then rotates every vector through an angle of π/6.
1√
1
cos (π/6) − sin (π/6)
1 0
2 3
2√
=
1
sin (π/6) cos (π/6)
0 −1
− 21 3
2
14. Find the matrix for the linear transformation which reflects every vector in R2 across
the y axis and then rotates every vector through an angle of π/6.
1√
cos (π/6) − sin (π/6)
−1 0
− 2 3 −√12
=
1
sin (π/6) cos (π/6)
0 1
− 12
2 3
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
13
Exercises
15. Find the matrix for the linear transformation which rotates every vector in R2 through
an angle of 5π/12. Hint: Note that 5π/12 = 2π/3 − π/4.
=
cos (2π/3) − sin (2π/3)
cos (−π/4) − sin (−π/4)
sin (2π/3) cos (2π/3)
sin (−π/4) cos (−π/4)
√
√ √
√ 1√ √
1
1
1
3 − 4√ 2
4 √2√3 − 4 √2 − 4√ 2
√
1
1
1
1
4 2 3+ 4 2
4 2 3− 4 2
16. Find the matrix for proju (v) where u = (1, −2, 3)T .
T ei =
ei ·(1,−2,3)
14
T
(1, −2, 3)
1
2
3
1
−2 −4 −6
14
3
6
9
17. Find the matrix for proju (v) where u = (1, 5, 3)T .
1
1
5
35
3
5 3
25 15
15 9
T
18. Find the matrix for proju (v) where u = (1, 0, 3) .
1 0
1
0 0
10
3 0
3
0
9
19. Give an example of a 2 × 2 matrix A which has all its entries nonzero and satisfies
A2 = A. Such a matrix is called idempotent.
You know it can’t be invertible. So try this.
2 2
a a
a + ba a2 + ba
=
b b
b2 + ab b2 + ab
Let a2 + ab = a, b2 + ab = b. A solution which yields a nonzero matrix is
2
2
−1 −1
20. Let A be an m × n matrix and let B be an n × m matrix where n < m. Show that
AB cannot have an inverse.
This follows right away from Theorem 2.3.8. This theorem says there exists a vector
x 6= 0 such that Bx = 0. Therefore, ABx = 0 also and so AB cannot be invertible.
21. Find ker (A) for
1
0
A=
1
0
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
2
2
4
2
3
1
4
1
2
1
3
1
1
2
.
3
2
7KH6D\ORU)RXQGDWLRQ
14
Exercises
Recall ker (A) is
1 2 3 2
0 2 1 1
1 4 4 3
0 2 1 1
just the set of solutions to Ax = 0.
1 0
1
0
2 0
. After row operations,
0
3 0
2 0
0
0
1
0
0
2
1
1
2
1
2
0
0
0
0
−1
1
0
0
0
0
0
0
A solution is x2 = − 12 t1 − 12 t2 − t3 , x1 = −2t1 − t2 + t3 where the ti are arbitrary.
22. If A is a linear transformation, and Axp = b. Show that the general solution to the
equation Ax = b is of the form xp + y where y ∈ ker (A). By this I mean to show
that whenever Az = b there exists y ∈ ker (A) such that xp + y = z.
If Az = b, Then A (z − xp ) = Az − Axp = b − b = 0 so there exists y such that
y ∈ ker (A) and xp + y = z.
23. Using Problem 21, find the general solution to the following linear system.
1
0
1
0
−2t1 − t2 + t3
− 21 t1 − 21 t2 − t3
t1
t2
t3
2
2
4
2
+
4
7/2
0
0
0
3
1
4
1
2
1
3
1
x1
1
x2
2
x3 =
3
x4
2
x5
11
7
18
7
, ti ∈ F
That second vector is a particular solution.
24. Using Problem 21, find the general solution to the following linear system.
1
0
1
0
−2t1 − t2 + t3
− 21 t1 − 21 t2 − t3
t1
t2
t3
2
2
4
2
+
−1
7/2
0
0
0
3
1
4
1
2
1
3
1
x1
1
x2
2
x3 =
3
x4
2
x5
6
7
13
7
, ti ∈ F
25. Show that the function Tu defined by Tu (v) ≡ v − proju (v) is also a linear transformation.
This is the sum of two linear transformations so it is obviously linear.
26. If u = (1, 2, 3)T and Tu
satisfies Au x = Tu (x).
1
1 0 0
0 1 0 − 1 2
14
3
0 0 1
is given in the above problem, find the matrix Au which
13
2 3
14
4 6 = − 17
3
6 9
− 14
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
− 17
5
7
− 37
3
− 14
− 37
5
14
7KH6D\ORU)RXQGDWLRQ
15
Exercises
27. ↑Suppose V is a subspace of Fn and T : V → Fp is a nonzero linear transformation.
Show that there exists a basis for Im (T ) ≡ T (V )
{T v1 , · · · , T vm }
and that in this situation,
{v1 , · · · , vm }
is linearly independent.
Im (T ) is a subspace of Fp . Therefore, it has a basis {T v1 , · · · , T vm } for some m ≤ p.
Say
m
X
ci vi = 0
i=1
Then do T to both sides
m
X
ci T vi = T 0 = 0
i=1
Hence each ci = 0. (T 0 =T (0 + 0) = T 0 + T 0 and so T 0 = 0)
28. ↑In the situation of Problem 27 where V is a subspace of Fn , show that there exists
{z1 , · · · , zr } a basis for ker (T ) . (Recall Theorem 2.4.12. Since ker (T ) is a subspace,
it has a basis.) Now for an arbitrary T v ∈ T (V ) , explain why
T v = a1 T v1 + · · · + am T vm
and why this implies
v − (a1 v1 + · · · + am vm ) ∈ ker (T ) .
Then explain why V = span (v1 , · · · , vm , z1 , · · · , zr ) .
ker (T ) is also a subspace so it has a basis {z1 , · · · , zr } for some r ≤ n. Now let the
basis for T (V ) be as above. Then for v an arbitrary vector, there exist unique scalars
ai such that
T v = a1 T v1 + · · · + am T vm
Then it follows that
T (v − (a1 v1 + · · · + am vm )) = T v− (a1 T v1 + · · · + am T vm ) = 0
Hence the vector v−(a1 v1 + · · · + am vm ) is in ker (T ) and so there exist unique scalars
bi such that
r
X
v − (a1 v1 + · · · + am vm ) =
bi zi
i=1
and therefore, V = span (v1 , · · · , vm , z1 , · · · , zr ) .
29. ↑In the situation of the above problem, show {v1 , · · · , vm , z1 , · · · , zr } is a basis for V
and therefore, dim (V ) = dim (ker (T )) + dim (T (V ))
The claim that {v1 , · · · , vm , z1 , · · · , zr } is a basis will be complete if it is shown that
these vectors are linearly independent. Suppose then that
X
X
ai zi +
bj vj = 0.
i
j
P
Then do T to both sides. Then you have j bj T vj = 0 and so each bj = 0. Then you
use the linear independence to conclude that each ai = 0.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
16
Exercises
30. ↑Let A be a linear transformation from V to W and let B be a linear transformation
from W to U where V, W, U are all subspaces of some Fp . Explain why
A (ker (BA)) ⊆ ker (B) , ker (A) ⊆ ker (BA) .
ker(BA)
ker(B)
A
ker(A)
-
A(ker(BA))
If x ∈ ker (BA) , then BAx = 0 and so Ax ∈ ker (B) . That is, BAx = 0. It follows
that
A (ker (BA)) ⊆ ker (B)
The second inclusion is obvious because if x is sent to 0 by A, then B will send Ax
to 0.
31. ↑Let {x1 , · · · , xn } be a basis of ker (A) and let {Ay1 , · · · , Aym } be a basis of A (ker (BA)).
Let z ∈ ker (BA) . Explain why
Az ∈ span {Ay1 , · · · , Aym }
and why there exist scalars ai such that
A (z − (a1 y1 + · · · + am ym )) = 0
and why it follows z − (a1 y1 + · · · + am ym ) ∈ span {x1 , · · · , xn }. Now explain why
ker (BA) ⊆ span {x1 , · · · , xn , y1 , · · · , ym }
and so
dim (ker (BA)) ≤ dim (ker (B)) + dim (ker (A)) .
This important inequality is due to Sylvester. Show that equality holds if and only if
A(ker BA) = ker(B).
Let {Ax1 , · · · , Axr } be a basis for A (ker (BA)) . Then let y ∈ ker (BA) . Thus It
follows that
Ay ∈ A (ker (BA)) .
Then there are scalars ai such that
Ay =
r
X
ai Axi
i=1
Hence y−
that
Pr
i=1
ai xi ∈ ker (A) . Let {z1 , · · · , zm } be a basis for ker (A) . This shows
dim ker (BA)
≤
≤
m + r = dim ker (A) + dim A (ker (BA))
dim ker (A) + dim ker (B)
It is interesting to note that the set of vectors {z1 , · · · , zm , x1 , · · · , xr } is independent.
To see this, say
X
X
aj zj +
bi xi = 0
j
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
i
7KH6D\ORU)RXQGDWLRQ
17
Exercises
Then do A to both sides and obtain each bi = 0. Then it follows that each aj is also
zero because of the independence of the zj . Thus
dim ker (BA) = dim ker (A) + dim (A ker (BA))
32. Generalize the result of the previous problem to any finite product of linear mappings.
Qs
Ps
dim (ker i=1 Ai ) ≤ i=1 dim ker (Ai ). This follows by induction.
33. If W ⊆ V for W, V two subspaces of Fn and if dim (W ) = dim (V ) , show W = V .
Let a basis for W be {w1 , · · · , wr } Then if there exists v ∈ V \ W, you could add in v
to the basis and obtain a linearly independent set of vectors of V which implies that
the dimension of V is at least r + 1 contrary to assumption.
34. Let V be a subspace of Fn and let V1 , · · · , Vm be subspaces, each contained in V . Then
V = V1 ⊕ · · · ⊕ Vm
(6.27)
if every v ∈ V can be written in a unique way in the form
v = v1 + · · · + vm
where each vi ∈ Vi . This is called a direct sum. If this uniqueness condition does not
hold, then one writes
V = V1 + · · · + Vm
and this symbol means all vectors of the form
v1 + · · · + vm , vj ∈ Vj for each j.
Show this is equivalent to saying that if
0 = v1 + · · · + vm , vj ∈ Vj for each j,
then each vj = 0. Next show that in the this situation, if β i = ui1 , · · · , uimi is a basis
for Vi , then {β 1 , · · · , β m } is a basis for V .
span (β 1 , · · · , β m ) is given to equal V. It only remains to verify that {β 1 , · · · , β m } is
linearly independent. Suppose vi ∈ Vi with
X
vi = 0
i
It is also true that
P
i
0 = 0 and so each vi = 0 by assumption. Now suppose
XX
i
cij uij = 0
j
Then from what was just observed, for each i,
X
cij uij = 0
j
and now, since these uij form a basis for Vi , it follows that each cij = 0 for each j for
each i. Thus {β 1 , · · · , β m } is a basis.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
18
Exercises
35. ↑Suppose you have finitely many linear mappings L1 , L2 , · · · , Lm which map V to V
where V is a subspace of Fn and suppose they commute. That is, Li Lj = Lj Li for all
i, j. Also suppose Lk is one to one on ker (Lj ) whenever j 6= k. Letting P denote the
product of these linear transformations, P = L1 L2 · · · Lm , first show
ker (L1 ) + · · · + ker (Lm ) ⊆ ker (P )
Next show Lj : ker (Li ) → ker (Li ) . Then show
ker (L1 ) + · · · + ker (Lm ) = ker (L1 ) ⊕ · · · ⊕ ker (Lm ) .
Using Sylvester’s theorem, and the result of Problem 33, show
ker (P ) = ker (L1 ) ⊕ · · · ⊕ ker (Lm )
Hint: By Sylvester’s theorem and the above problem,
X
dim (ker (P )) ≤
dim (ker (Li ))
i
= dim (ker (L1 ) ⊕ · · · ⊕ ker (Lm )) ≤ dim (ker (P ))
Now consider Problem 33.
First note that, since these operators
commute, it follows that Lk : ker Li → ker Li .
P
Let vi ∈ ker (Li ) P
and consider i vi . It is obvious, since the linear transformations
commute that P ( i vi ) = 0. Thus
ker (L1 ) + · · · + ker (Lm ) ⊆ ker (P )
However, by Sylvester’s theorem
dim ker (P ) ≤
If vi ∈ ker (Li ) and
P
i
vi = 0, then apply
Y
X
dim ker (Li )
i
Q
i6=k
Li to both sides. This yields
Li vk = 0
i6=k
Since each of these Li is one to one on ker (Lk ) , it follows that vk = 0. Thus
ker (L1 ) + · · · + ker (Lm ) = ker (L1 ) ⊕ · · · ⊕ ker (Lm )
P
Now it follows that a basis for ker (L1 ) + · · · + ker (Lm ) has i dim ker (Li ) vectors in
it.
X
dim ker (P ) ≤
dim ker (Li )
i
=
≤
dim (ker (L1 ) + · · · + ker (Lm ))
dim ker (P )
Thus the inequalities are all equal signs and so
ker (L1 ) ⊕ · · · ⊕ ker (Lm ) = ker (P )
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
19
Exercises
36. Let M (Fn , Fn ) denote the set of all n × n matrices having entries in F. With the usual
operations of matrix addition and scalar multiplications, explain why M (Fn , Fn ) can
2
be considered as Fn . Give a basis for M (Fn , Fn ) . If A ∈ M (Fn , Fn ) , explain why
there exists a monic polynomial of the form
λk + ak λk + · · · + a1 λ + a0
such that
Ak + ak Ak + · · · + a1 A + a0 I = 0
The minimal polynomial of A is the polynomial like the above, for which p (A) = 0
which has smallest degree. I will discuss the uniqueness of this polynomial later. Hint:
2
Consider the matrices I, A, A2 , · · · , An . There are n2 + 1 of these matrices. Can they
be linearly independent? Now consider all polynomials and pick one of smallest degree
and then divide by the leading coefficient.
A basis for M (Fn , Fn ) is obviously the matricies Eij where Eij has a 1 in the ij th
place and zeros everywhere else. Thus the dimension of this vector space is n2 . It
2
follows that the list of matrices I, A, A2 , · · · , An is linearly dependent. Hence there
exists a polynomial p (λ) which has smallest possible degree such that p (A) = 0 by the
well ordering principle of the natural numbers. Then divide by the leading coefficient.
If you insist that its leading coefficient be 1, (monic) then the polynomial is unique
and it is called the minimal polynomial. It is unique thanks to the division algorithm,
because if q (λ) is another one, then
q (λ) = p (λ) l (λ) + r (λ)
where the degree of r (λ) is less than the degree of p (λ) or else equals 0. If it is not
zero, then r (A) = 0 and this would be a contradiction. Hence q (λ) = p (λ) l (λ) where
l (λ) must be monic. Since q (λ) has smallest possible degree, this monic polynomial
can only be 1. Thus q (λ) = p (λ).
37. ↑Suppose the field of scalars is C and A is an n × n matrix. From the preceding
problem, and the fundamental theorem of algebra, this minimal polynomial factors
(λ − λ1 )r1 (λ − λ2 )r2 · · · (λ − λk )rk
where rj is the algebraic multiplicity of λj . Thus
r
r
rk
(A − λ1 I) 1 (A − λ2 I) 2 · · · (A − λk I)
=0
and so, letting P = (A − λ1 I)r1 (A − λ2 I)r2 · · · (A − λk I)rk and Lj = (A − λj I)rj
apply the result of Problem 35 to verify that
Cn = ker (L1 ) ⊕ · · · ⊕ ker (Lk )
and that A : ker (Lj ) → ker (Lj ). In this context, ker (Lj ) is called the generalized
eigenspace for λj . You need to verify the conditions of the result of this problem hold.
r
Let Li = (A − λi I) i . Then obviously these commute since they are just polynomials
in A. Is Lk one to one on ker (Li )?
rk
(A − λk I)
rk
= (A − λi I + (λi − λk ) I)
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
20
Exercises
rk X
rk
j
r −j
(A − λi I) (λi − λk ) k
j
j=0
rk X
rk
rk
= (λi − λk ) I +
(A − λi I)j (λi − λk )rk −j
j
=
j=1
Now raise both sides to the ri power.
(A − λk I)
rk ri
rk ri
= (λi − λk )
ri
I + g (A) (A − λi I)
where g (A) is some polynomial in A. Let vi ∈ ker (Li ) . Then suppose (A − λk I)rk vi =
0. Then
r r
r r
0 = (A − λk I) k i vi = (λi − λk ) k i vi
and since λi 6= λk , this requires that
Q vi = 0. Thus Lk is one to one on ker (Li ) as
hoped. Therefore, since Cn = ker i Li , it follows from the above problems that
Cn = ker (L1 ) ⊕ · · · ⊕ ker (Lk )
Note that there was nothing sacred about C all you needed for the above to hold is
that the minimal polynomial factors completely into a product of linear factors. In
other words, all the above works fine for Fn provided the minimal polynomial “splits”.
38. In the context of Problem 37, show there exists a nonzero vector x such that
(A − λj I) x = 0.
This is called an eigenvector and the λj is called an eigenvalue. Hint: There must
exist a vector y such that
rj −1
r
r
(A − λ1 I) 1 (A − λ2 I) 2 · · · (A − λj I)
· · · (A − λk I)
rk
y = z 6= 0
Why? Now what happens if you do (A − λj I) to z?
The hint gives it away.
r −1
r
r
r
(A − λj I) z = (A − λj I) (A − λ1 I) 1 (A − λ2 I) 2 · · · (A − λj I) j · · · (A − λk I) k y
= (A − λ1 I)r1 (A − λ2 I)r2 · · · (A − λj I)rj · · · (A − λk I)rk y = 0
39. Suppose Q (t) is an orthogonal matrix. This means Q (t) is a real n × n matrix which
satisfies
T
Q (t) Q (t) = I
0
Suppose also the entries of Q (t) are differentiable. Show QT = −QT Q0 QT .
This is just the product rule.
T
T
Q0 (t) Q (t) + Q (t) Q0 (t) = 0
Hence
T
T
Q0 (t) = −Q (t) Q0 (t) Q (t)
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
T
7KH6D\ORU)RXQGDWLRQ
21
Exercises
40. Remember the Coriolis force was 2Ω × vB where Ω was a particular vector which
came from the matrix Q (t) as described above. Show that
i (t) · i (t0 ) j (t) · i (t0 ) k (t) · i (t0 )
Q (t) = i (t) · j (t0 ) j (t) · j (t0 ) k (t) · j (t0 ) .
i (t) · k (t0 ) j (t) · k (t0 ) k (t) · k (t0 )
There will be no Coriolis force exactly when Ω = 0 which corresponds to Q0 (t) = 0.
When will Q0 (t) = 0?
Recall, that letting i = e1 , j = e2 , k = e3 in the usual way,
Q (t) u = u1 e1 (t) + u2 e2 (t) + u3 e3 (t)
where
u ≡ u1 e1 (t0 ) + u2 e2 (t0 ) + u3 e3 (t0 )
Note that uj = u · ej (t0 ) . Thus
Q (t) u =
X
j
u · ej (t0 ) ej (t)
So what is the rsth entry of Q (t)? It equals
T
er (t0 ) Q (t) es (t0 )
= er (t0 ) ·
X
j
es (t0 ) ·ej (t0 ) ej (t)
= er (t0 ) · es (t)
which shows the desired result.
41. An illustration used in many beginning physics books is that of firing a rifle horizontally and dropping an identical bullet from the same height above the perfectly
flat ground followed by an assertion that the two bullets will hit the ground at exactly the same time. Is this true on the rotating earth assuming the experiment
takes place over a large perfectly flat field so the curvature of the earth is not an
issue? Explain. What other irregularities will occur? Recall the Coriolis acceleration
is 2ω [(−y 0 cos φ) i+ (x0 cos φ + z 0 sin φ) j − (y 0 sin φ) k] where k points away from the
center of the earth, j points East, and i points South.
Obviously not. Because of the Coriolis force experienced by the fired bullet which is
not experienced by the dropped bullet, it will not be as simple as in the physics books.
For example, if the bullet is fired East, then y 0 sin φ > 0 and will contribute to a force
acting on the bullet which has been fired which will cause it to hit the ground faster
than the one dropped. Of course at the North pole or the South pole, things should
be closer to what is expected in the physics books because there sin φ = 0. Also, if
you fire it North or South, there seems to be no extra force because y 0 = 0.
F.18
Exercises
3.2
1. Find the determinants of the following matrices.
1 2 3
(a) 3 2 2 (The answer is 31.)
0 9 8
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
22
Exercises
4
(b) 1
3
1
1
(c)
4
1
3 2
7 8 (The answer is 375.)
−9 3
2 3 2
3 2 3
, (The answer is −2.)
1 5 0
2 1 2
2. If A−1 exist, what is the relationship between det (A) and det A−1 . Explain your
answer.
1 = det AA−1 = det (A) det A−1 .
3. Let A be an n × n matrix where n is odd. Suppose also that A is skew symmetric.
This means AT = −A. Show that det(A) = 0.
det (A) = det AT = det (−A) = det (−I) det (A) = (−1)n det (A) = − det (A) .
4. Is it true that det (A + B) = det (A) + det (B)? If this is so, explain why it is so and
if it is not so, give a counter example.
Almost anything shows that this is not true.
1 0
−1 0
det
+
=
0 1
0 −1
1 0
−1 0
det
+ det
=
0 1
0 −1
0
2
5. Let A be an r × r matrix and suppose there are r − 1 rows (columns) such that all rows
(columns) are linear combinations of these r − 1 rows (columns). Show det (A) = 0.
Without loss of generality, assume the last row is a linear combination of the first r − 1
rows. Then the matrix is of the form
rT1
..
.
rTn−1
Pn−1
T
a
r
i=1 i i
Then from the linear property of determinants, the determinant equals
T
T
r1
r1
n−1
.. n−1
..
X
X
ai det . =
ai det . = 0
T
T
rn−1
rn−1
i=1
i=1
T
ri
0T
Where the first equal sign in the above is obtained by taking −1 times a the ith row
from the top and adding to the last row.
6. Show det (aA) = an det (A) where here A is an n × n matrix and a is a scalar.
Each time you take out an a from a row, you multiply by a the determinant of the
matrix which remains. Since there are n rows, you do this n times, hence you get an .
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
23
Exercises
7. Suppose A is an upper triangular matrix. Show that A−1 exists if and only if all
elements of the main diagonal are non zero. Is it true that A−1 will also be upper
triangular? Explain. Is everything the same for lower triangular matrices?
This is obvious because the determinant of A is the product of these diagonal entries.
When you consider the usual process of finding the inverse, you get that A−1 must be
upper triangular. Everything is similar for lower triangular matrices.
8. Let A and B be two n × n matrices. A ∼ B (A is similar to B) means there exists an
invertible matrix S such that A = S −1 BS. Show that if A ∼ B, then B ∼ A. Show
also that A ∼ A and that if A ∼ B and B ∼ C, then A ∼ C.
This is easy except possibly for the last claim. Say A = P −1 BP and B = Q−1 CQ.
Then
A = P −1 BP = A = P −1 Q−1 CQP = (QP )−1 C (QP )
9. In the context of Problem 8 show that if A ∼ B, then det (A) = det (B) .
det A =
=
det P −1 BP = det P −1 det (B) det (P )
det (B) det P −1 P = det (B) .
10. Let A be an n × n matrix and let x be a nonzero vector such that Ax = λx for some
scalar, λ. When this occurs, the vector, x is called an eigenvector and the scalar, λ
is called an eigenvalue. It turns out that not every number is an eigenvalue. Only
certain ones are. Why? Hint: Show that if Ax = λx, then (λI − A) x = 0. Explain
why this shows that (λI − A) is not one to one and not onto. Now use Theorem 3.1.15
to argue det (λI − A) = 0. What sort of equation is this? How many solutions does it
have?
−1
If you have (λI − A) x = 0 for x 6= 0, then (λI − A) cannot exist because if it did,
you could multiply on the left by it and then conclude that x = 0. Therefore, (λI − A)
is not one to one and not onto.
11. Suppose det (λI − A) = 0. Show using Theorem 3.1.15 there exists x 6= 0 such that
(λI − A) x = 0.
If that determinant equals 0 then the matrix λI − A has no inverse. It is not one
to one and so there exists x 6= 0 such that (λI − A) x = 0. Also recall the process for
finding the inverse.
a (t) b (t)
12. Let F (t) = det
. Verify
c (t) d (t)
0
F (t) = det
Now suppose
a0 (t) b0 (t)
c (t) d (t)
+ det
a (t) b (t)
c0 (t) d0 (t)
.
a (t) b (t) c (t)
F (t) = det d (t) e (t) f (t) .
g (t) h (t) i (t)
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
24
Exercises
Use Laplace expansion and the first part to verify F 0 (t) =
0
a (t) b (t) c (t)
a (t) b0 (t) c0 (t)
det d (t) e (t) f (t) + det d0 (t) e0 (t) f 0 (t)
g (t) h (t) i (t)
g (t) h (t) i (t)
a (t) b (t) c (t)
+ det d (t) e (t) f (t) .
g 0 (t) h0 (t) i0 (t)
Conjecture a general result valid for n × n matrices and explain why it will be true.
Can a similar thing be done with the columns?
The way to see this holds in general is to use the usual proof for the product rule and
the theorem about the determinant and row operations.
a (t + h) b (t + h) c (t + h)
F (t + h) − F (t) = det d (t + h) e (t + h) f (t + h)
g (t + h) h (t + h) i (t + h)
a (t) b (t) c (t)
− det d (t) e (t) f (t)
g (t) h (t) i (t)
And so this equals
a (t + h) b (t + h) c (t + h)
a (t)
b (t)
c (t)
det d (t + h) e (t + h) f (t + h) − det d (t + h) e (t + h) f (t + h)
g (t + h) h (t + h) i (t + h)
g (t + h) h (t + h) i (t + h)
a (t)
b (t)
c (t)
a (t)
b (t)
c (t)
e (t)
f (t)
+ det d (t + h) e (t + h) f (t + h) − det d (t)
g (t + h) h (t + h) i (t + h)
g (t + h) h (t + h) i (t + h)
a (t)
b (t)
c (t)
a (t) b (t) c (t)
e (t)
f (t) − det d (t) e (t) f (t)
+ det d (t)
g (t + h) h (t + h) i (t + h)
g (t) h (t) i (t)
Now multiply by 1/h to obtain the following for the difference quotient
det
a(t+h)−a(t)
h
d (t + h)
g (t + h)
b(t+h)−b(t)
h
e (t + h)
h (t + h)
+ det
c(t+h)−c(t)
h
f (t + h) +det
i (t + h)
F (t+h)−F (t)
.
h
a (t)
b (t)
c (t)
d(t+h)−d(t)
h
e(t+h)−e(t)
h
f (t+h)−f (t)
h
g (t + h)
a (t)
d (t)
b (t)
e (t)
c (t)
f (t)
g(t+h)−g(t)
h
h(t+h)−h(t)
h
i(t+h)−i(t)
h
h (t + h)
i (t + h)
Now passing to a limit yields the desired formula. Obviously this holds for any size
determinant.
13. Use the formula for the inverse in terms of the cofactor matrix to find the inverse of
the matrix
t
e
0
0
.
et cos t
et sin t
A= 0
t
t
t
0 e cos t − e sin t e cos t + et sin t
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
25
Exercises
−1
0
0
et cos t
et sin t
t
t
t
t
e cos t − e sin t e cos t + e sin t
−t
e
0
0
e−t (cos t + sin t) − (sin t) e−t
= 0
0 −e−t (cos t − sin t) (cos t) e−t
et
0
0
14. Let A be an r × r matrix and let B be an m × m matrix such that r + m = n. Consider
the following n × n block matrix
A 0
C=
.
D B
where the D is an m × r matrix, and the 0 is a r × m matrix. Letting Ik denote the
k × k identity matrix, tell why
A 0
Ir 0
C=
.
D Im
0 B
Now explain why det (C) = det (A) det (B) . Hint: Part of this will require an explanation of why
A 0
det
= det (A) .
D Im
See Corollary 3.1.9.
The first follows right away from block multiplication. Now
A 0
Ir 0
det (C) = det
det
D Im
0 B
A 0
Ir 0
= det
det
= det (A) det (B)
0 Im
0 B
from expanding along the last m columns for the first one and along the first r columns
for the second.
15. Suppose Q is an orthogonal matrix. This means Q is a real n×n matrix which satisfies
QQT = I
Find the possible values for det (Q).
2
You have to have det (Q) det QT = det (Q) = 1 and so det (Q) = ±1.
16. Suppose Q (t) is an orthogonal matrix. This means Q (t) is a real n × n matrix which
satisfies
Q (t) Q (t)T = I
Suppose Q (t) is continuous for t ∈ [a, b] , some interval. Also suppose det (Q (t)) = 1.
Show that it follows det (Q (t)) = 1 for all t ∈ [a, b].
You have from the given equation that det (Q (t)) is always either 1 or −1. Since Q (t)
is continuous, so is t → det (Q (t)) and so if it starts off at 1, it cannot jump to −1
because this would violate the intermediate value theorem.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
26
Exercises
F.19
Exercises
3.6
1. Let m < n and let A be an m × n matrix. Show that A is not one to one. Hint:
Consider the n × n matrix A1 which is of the form
A
A1 ≡
0
where the 0 denotes an (n − m) × n matrix of zeros. Thus det A1 = 0 and so A1 is
not one to one. Now observe that A1 x is the vector,
Ax
A1 x =
0
which equals zero if and only if Ax = 0.
The hint gives it away. You could simply consider a vector of the form
0
a
where a 6= 0.
2. Let v1 , · · · , vn be vectors in Fn and let M (v1 , · · · , vn ) denote the matrix whose ith
column equals vi . Define
d (v1 , · · · , vn ) ≡ det (M (v1 , · · · , vn )) .
Prove that d is linear in each variable, (multilinear), that
d (v1 , · · · , vi , · · · , vj , · · · , vn ) = −d (v1 , · · · , vj , · · · , vi , · · · , vn ) ,
(6.28)
and
d (e1 , · · · , en ) = 1
(6.29)
where here ej is the vector in Fn which has a zero in every position except the j th
position in which it has a one.
This follows from the properties of determinants which are discussed above.
3. Suppose f : Fn × · · · × Fn → F satisfies 6.28 and 6.29 and is linear in each variable.
Show that f = d.
Consider f (x1 , · · · , xn ) . Then by the assumptions on f it equals
X
f (x1 , · · · , xn ) =
xi1 · · · xin f (ei1 · · · ein )
i1 ,··· ,in
=
X
i1 ,··· ,in
=
X
i1 ,··· ,in
=
X
i1 ,··· ,in
x1i1 · · · x1in sgn (i1 , · · · , in ) f (e1 · · · e1 )
x1i1 · · · x1in sgn (i1 , · · · , in ) d (e1 · · · e1 )
x1i1 · · · x1in d (ei1 · · · ein ) = d (x1 , · · · , xn )
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
27
Exercises
4. Show that if you replace a row (column) of an n × n matrix A with itself added to
some multiple of another row (column) then the new matrix has the same determinant
as the original one.
This was done above.
5. Use the result of Problem 4 to evaluate by hand
1 2 3
−6 3 2
det
5 2 2
3 4 6
1
−6
det
5
3
2
3
2
4
3
2
2
6
2
3
=5
3
4
6. Find the inverse if it exists of the matrix
t
e
cos t
et − sin t
et − cos t
et
et
et
=
cos t
− sin t
− cos t
1
2
1
2
the determinant
2
3
.
3
4
−1
sin t
cos t
− sin t
1 −t
2e
cos t + 12 sin t
sin t − 12 cos t
0
− sin t
cos t
sin t
cos t .
− sin t
1 −t
2e
1
1
2 sin t − 2 cos t
1
− 2 cos t − 12 sin t
7. Let Ly = y (n) + an−1 (x) y (n−1) + · · · + a1 (x) y 0 + a0 (x) y where the ai are given
continuous functions defined on a closed interval, (a, b) and y is some function which
has n derivatives so it makes sense to write Ly. Suppose Lyk = 0 for k = 1, 2, · · · , n.
The Wronskian of these functions, yi is defined as
y1 (x)
···
yn (x)
0
0
y1 (x)
···
yn (x)
W (y1 , · · · , yn ) (x) ≡ det
..
..
.
.
(n−1)
y1
(x)
···
(n−1)
yn
(x)
Show that for W (x) = W (y1 , · · · , yn ) (x) to save space,
W 0 (x) = det
y1 (x)
y10 (x)
..
.
(n)
···
···
y1 (x) · · ·
yn (x)
yn0 (x)
..
.
(n)
yn (x)
.
Now use the differential equation, Ly = 0 which is satisfied by each of these functions,
yi and properties of determinants presented above to verify that W 0 + an−1 (x) W = 0.
Give an explicit solution of this linear differential equation, Abel’s formula, and use
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
28
Exercises
your answer to verify that the Wronskian of these solutions to the equation, Ly = 0
either vanishes identically on (a, b) or never.
The last formula above follows because W 0 (x) equals the sum of determinants of
matrices which have two equal rows except for the last one in the sum which is the
displayed expression. Now let
(n−1)
mi (x) = − an−1 (x) yi
+ · · · + a1 (x) yi0 + a0 (x) yi
(n)
Since each yi is a solution to Ly = 0, it follows that yi (t) = mi (t). Now from the
properties of determinants, being linear in each row,
y1 (x)
···
yn (x)
y10 (x)
···
yn0 (x)
0
W (x) = −an−1 (x) det
..
..
.
.
(n−1)
(n−1)
y1
(x) · · · yn
(x)
=
−an−1 (x) W 0 (x)
Now let A0 (x) = an−1 (x) . Then
d A(x)
e
W (x) = 0
dx
and so W (x) = Ce−A(x) . Thus the Wronskian either vanishes for all x or for no x.
8. Two n × n matrices, A and B, are similar if B = S −1 AS for some invertible n × n
matrix S. Show that if two matrices are similar, they have the same characteristic
polynomials. The characteristic polynomial of A is det (λI − A) .
Say A = S −1 BS. Then
det (λI − A)
det λI − S −1 BS
=
det λS −1 S − S −1 BS
det S −1 (λI − B) S
det S −1 det (λI − B) det (S)
det S −1 S det (λI − B) = det (λI − B)
=
=
=
=
9. Suppose the characteristic polynomial of an n × n matrix A is of the form
tn + an−1 tn−1 + · · · + a1 t + a0
and that a0 6= 0. Find a formula A−1 in terms of powers of the matrix A. Show that
n
A−1 exists if and only if a0 6= 0.In fact a0 = (−1) det (A).
From the Cayley Hamilton theorem,
An + an−1 An−1 + · · · + a1 A + a0 I = 0
Also the characteristic polynomial is
n
det (tI − A)
and the constant term is (−1) det (A) . Thus a0 6= 0 if and only if det (A) 6= 0 if and
only if A−1 has an inverse. Thus if A−1 exists, it follows that
a0 I = − An + an−1 An−1 + · · · + a1 A = A −An−1 − an−1 An−2 − · · · − a1 I
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
29
Exercises
and also
a0 I = −An−1 − an−1 An−2 − · · · − a1 I A
Therefore, the inverse is
1
−An−1 − an−1 An−2 − · · · − a1 I
a0
10. ↑Letting p (t) denote the characteristic polynomial of A, show that
pε (t) ≡ p (t − ε)
is the characteristic polynomial of A + εI. Then show that if det (A) = 0, it follows
that det (A + εI) 6= 0 whenever |ε| is sufficiently small.
p (t) ≡ det (tI − A) . Hence p (t − ε) = det ((t − ε) I − A) = det (tI − (εI + A)) and
by definition, this is the characteristic polynomial of εI + A. Therefore, the constant
term of this is the constant term of p (t − ε). Say
p (t) = tn + an−1 tn−1 + · · · + a1 t + a0
where the smallest ai which is nonzero is ak . Then it reduces to
tn + an−1 tn−1 + · · · + ak tk
Then
n
n−1
pε (t) = p (t − ε) = (t − ε) + an−1 (t − ε)
k
+ · · · + ak (t − ε)
k
then the constant term of this pε (t) is ak (−1) εk + terms having ε raised to a higher
power. Hence for all ε small enough, this term will dominate the sum of all the others
and it follows that the constant term is nonzero for all |ε| small enough.
11. In constitutive
modeling of the stress and strain tensors, one sometimes considers sums
P∞
of the form k=0 ak Ak where A is a 3×3 matrix. Show using the Cayley Hamilton
theorem that if such a thing makes any sense, you can always obtain it as a finite sum
having no more than n terms.
Say the characteristic polynomial is q (t) . Then if n ≥ 3,
tn = q (t) l (t) + r (t)
where the degree of r (t) is either less than 3 or it equals zero. Thus
An = q (A) l (A) + r (A) = r (A)
and so all the terms An for n ≥ 3 can be replaced with some r (A) where the degree of
r (t) is no more than 2. Thus, assuming there are no convergence issues, the infinite
P2
sum must be of the form k=0 bk Ak .
12. Recall you can find the determinant from expanding along the j th column.
X
det (A) =
Aij (cof (A))ij
i
Think of det (A) as a function of the entries, Aij . Explain why the ij th cofactor is
really just
∂ det (A)
.
∂Aij
This follows from the formula for determinant. If you take the partial derivative, you
get (cof (A))ij .
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
30
Exercises
13. Let U be an open set in Rn and let g :U → Rn be such that all the first partial
derivatives of all components of g exist and are continuous. Under these conditions
form the matrix Dg (x) given by
Dg (x)ij ≡
∂gi (x)
≡ gi,j (x)
∂xj
The best kept secret in calculus courses is that the linear transformation determined
by this matrix Dg (x) is called the derivative of g and is the correct generalization
of the concept of derivative of a function of one variable. Suppose the second partial
derivatives also exist and are continuous. Then show that
X
(cof (Dg))ij,j = 0.
j
Hint: First explain why
X
gi,k cof (Dg)ij = δ jk det (Dg)
i
Next differentiate with respect to xj and sum on j using the equality of mixed partial
derivatives. Assume det (Dg) 6= 0 to prove the identity in this special case. Then
explain why there exists a sequence εk → 0 such that for gεk (x) ≡ g (x) + εk x,
det (Dgεk ) 6= 0 and so the identity holds for gεk . Then take a limit to get the desired result in general. This is an extremely important identity which has surprising
implications.
First assume det (Dg) 6= 0. Then from the cofactor expansion, you get
X
gi,k cof (Dg)ij = δ jk det (Dg)
i
If j = k you get det (Dg) on the right but if j 6= k, then the left is the expansion of a
determinant which has two equal columns. Now differentiate both sides with respect
to j using the above problem. You have to use the chain rule.
X
X
gi,kj cof (Dg)ij +
gi,k cof (Dg)ij,j
i
= δ jk
i
X ∂ (det (Dg))
r,s
gr,s
gr,sj =
X
δ jk cof (Dg)rs gr,sj
r,s
Next sum on j
X
X
X
gi,kj cof (Dg)ij +
gi,k cof (Dg)ij,j =
δ jk cof (Dg)rs gr,sj
i,j
i,j
r,s,j
Of course the terms on the right are 0 unless j = k, and so
X
X
X
gi,kj cof (Dg)ij +
gi,k cof (Dg)ij,j =
cof (Dg)rs gr,sk
r,s
i,j
i,j
=
X
cof (Dg)ij gi,jk
i,j
Subtract the terms at the end from each side using equality of mixed partial derivatives.
This gives.
X
X
X
0=
gi,k cof (Dg)
=
gi,k
cof (Dg)
ij,j
i,j
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
ij,j
i
j
7KH6D\ORU)RXQGDWLRQ
31
Exercises
Since Dg is assumed invertible, this requires
X
cof (Dg)ij,j = 0
j
In case det (Dg (x)) = 0 for some x, consider εk , a sequence of numbers converging to
0 with the property that det (Dg (x) + εk I) 6= 0. Then let gk (x) = g (x) + εk x. From
the above, you have
X
X
cof (Dg)ij,j (x) = lim
cof (Dgk )ij,j (x) = 0
k→∞
j
j
14. A determinant of the form
1
a0
a20
..
.
n−1
a
0
an
0
1
a1
a21
..
.
···
···
···
1
an
a2n
..
.
an−1
1
an1
···
···
an−1
n
ann
is called a Vandermonde determinant. Show this determinant equals
Y
(aj − ai )
0≤i<j≤n
By this is meant to take the product of all terms of the form (aj − ai ) such that j > i.
Hint: Show it works if n = 1 so you are looking at
1 1 a0 a1 Then suppose it holds for n − 1 and consider the
polynomial.
1
1
···
a0
a
···
1
2
a20
a
···
1
p (t) ≡ .
..
.
.
.
n−1
n−1
a
a
···
1
0
n
an
a
···
0
1
Explain why p (aj ) = 0 for i = 0, · · · , n − 1. Thus
p (t) = c
n−1
Y
i=0
case n. Consider the following
1
t
t2
..
.
tn−1
tn
.
(t − ai ) .
Of course c is the coefficient of tn . Find this coefficient from the above description of
p (t) and the induction hypothesis. Then plug in t = an and observe you have the
formula valid for n.
1 1 = (a1 − a0 ) so the formula holds. Now
In the case of n = 1, you have a0 a1 suppose it holds for n and consider p (t) as above. Then it is obvious that p (aj ) = 0
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
32
Exercises
for i = 0, · · · , n − 1 because you have the determinant of a matrix with two equal
columns. Thus
n−1
Y
p (t) = c
(t − ai ) .
i=0
because p (t) has degree n and so it has no more than n roots. So what is c? The
n
coefficient
it
Q of t is the Vandermonde determinant which is n × n and by induction,
equals 0≤i<j≤n−1 (aj − ai ) . From the above product, the coefficient of tn is c. Thus
p (t) =
Y
0≤i<j≤n−1
(aj − ai )
n−1
Y
i=0
(t − ai )
Now plug in t = an to get the (n + 1) × (n + 1) Vandermonde determinant
Y
p (an ) =
0≤i<j≤n−1
=
Y
0≤i<j≤n
F.20
(aj − ai )
n−1
Y
i=0
(an − ai )
(aj − ai )
Exercises
4.6
1. Let {u1 , · · · , un } be vectors in Rn . The parallelepiped determined by these vectors
P (u1 , · · · , un ) is defined as
( n
)
X
P (u1 , · · · , un ) ≡
tk uk : tk ∈ [0, 1] for all k .
k=1
Now let A be an n × n matrix. Show that
{Ax : x ∈ P (u1 , · · · , un )}
is also a parallelepiped.
A typical thing in {Ax : x ∈ P (u1 , · · · , un )} is
n
X
k=1
tk Auk : tk ∈ [0, 1]
and so it is just P (Au1 , · · · , Aun ) .
2. In the context of Problem 1, draw P (e1 , e2 ) where e1 , e2 are the standard basis vectors
for R2 . Thus e1 = (1, 0) , e2 = (0, 1) . Now suppose
1 1
E=
0 1
where E is the elementary matrix which takes the third row and adds to the first.
Draw
{Ex : x ∈ P (e1 , e2 )} .
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
33
Exercises
In other words, draw the result of doing E to the vectors in P (e1 , e2 ). Next draw the
results of doing the other elementary matrices to P (e1 , e2 ).
For E the given elementary matrix, the result is as follows.
P (e1 , e2 )
E(P (e1 , e2 ))
It is called a shear. Note that it does not change the area.
1 0
In caseE =
, the given square P (e1 , e2 ) becomes a rectangle. If α > 0,
0 α
the square is stretched by a factor of α in the y direction. If α < 0 the rectangle
is reflected across the x axis and
in addition is stretched by a factor of |α| in the y
α 0
direction. When E =
the result is similar only it features changes in the x
0 1
direction. These elementary matrices
the area unless |α| = 1 when the area
do change
0 1
is unchanged. The permutation
simply switches e1 and e2 so the result
1 0
appears to be a square just like the one you began with.
3. In the context of Problem 1, either draw or describe the result of doing elementary
matrices to P (e1 , e2 , e3 ). Describe geometrically the conclusion of Corollary 4.3.7.
It is the same sort of thing. The elementary matrix either switches the ei about or it
produces a shear or a magnification in one direction.
4. Consider a permutation of {1, 2, · · · , n}. This is an ordered list of numbers taken from
this list with no repeats, {i1 , i2 , · · · , in }. Define the permutation matrix P (i1 , i2 , · · · , in )
as the matrix which is obtained from the identity matrix by placing the j th column
of I as the ith
j column of P (i1 , i2 , · · · , in ) . What does this permutation matrix do to
T
the column vector (1, 2, · · · , n) ?
T
It produces the column (i1 , i2 , · · · , in ) .
5. Consider the 3 × 3 permutation matrices. List all of them and then determine the
dimension of their span. Recall that you can consider an m × n matrix as something
in Fnm .
Here they are.
1
0
0
0
1
0
0 0
1 0 ,
0 1
1 0
0 0 ,
0 1
0 0
1 0
0 1
1 0
0 0
0 1
1
0 ,
0
0
1 ,
0
0
0
1
0
0
1
1 0
0 1
0 0
0 1
1 0
0 0
So what is the dimension of the span of these? One way to systematically accomplish
this is to unravel them and then use the row reduced echelon form. Unraveling these
yields the column vectors
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
34
Exercises
1
0
0
0
1
0
0
0
1
0
0
1
1
0
0
0
1
0
0
1
0
0
0
1
1
0
0
0
1
0
1
0
0
0
0
1
1
0
0
0
0
1
0
1
0
0
0
1
0
1
0
1
0
0
Then arranging these as the columns of a matrix
row reduced echelon form.
1 0 0 0 1 0
1
0 0 1 1 0 0
0
0 1 0 0 0 1
0
0 1 0 1 0 0
0
1 0 0 0 0 1 , row echelon form: 0
0 0 1 0 1 0
0
0 0 1 0 0 1
0
0 1 0 0 1 0
0
1 0 0 1 0 0
0
yields the following along with its
0
1
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
The dimension is 5.
0 1
0 1
0 1
0 −1
1 −1
0 0
0 0
0 0
0 0
6. Determine which matrices are in row reduced echelon form.
1 2 0
(a)
0 1 7
This one is not.
1 0 0 0
(b) 0 0 1 2
0 0 0 0
This one is.
1 1 0 0 0 5
(c) 0 0 1 2 0 4
0 0 0 0 1 3
This one is.
7. Row reduce the following matrices to obtain the row reduced echelon form. List the
pivot columns in the original matrix.
1 0 0 3
1 2 0 3
(a) 2 1 2 2 , row echelon form: 0 1 0 0
0 0 1 −2
1 1 0 3
1 2 3
1 0 0
0 1 0
2 1 −2
(b)
3 0 0 , row echelon form: 0 0 1
3 2 1
0 0 0
1 0 0 0
1 2 1 3
(c) −3 2 1 0 , row echelon form: 0 1 12 0
3 2 1 1
0 0 0 1
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
Exercises
35
8. Find the rank and nullity of the following matrices. If the rank is r, identify r columns
in the original matrix which have the property that every other column may be
written as a linear combination of these.
0 1 0 2 1 2 2
0 1 0 2 0 1 1
0 3 2 12 1 6 8
0 0 1 3 0 1 2
(a)
0 1 1 5 0 2 3 , row echelon form: 0 0 0 0 1 1 1 Rank
0 2 1 7 0 3 4
0 0 0 0 0 0 0
equals 3 nullity equals 4. A basis is columns 2,3,5
0 1 0 2 0 1 0
0 1 0 2 0 1 0
0 0 1 0 0 1 2
0 3 2 6 0 5 4
(b)
0 1 1 2 0 2 2 , row echelon form: 0 0 0 0 0 0 0 Rank
0 0 0 0 0 0 0
0 2 1 4 0 3 2
is 2 nullity equals 5. A basis is columns 2,3.
0 1 0 2 1 1 2
0 1 0 2 0 1 0
0 3 2 6 1 5 1
0 0 1 0 0 1 0
(c)
0 1 1 2 0 2 1 , row echelon form: 0 0 0 0 1 0 0 Rank
0 2 1 4 0 3 1
0 0 0 0 0 0 1
is 4 and nullity is 3. A basis is columns 2,3,5,7.
9. Find the rank of the following matrices. If the rank is r, identify r columns in the
original matrix which have the property that every other column may be written
as a linear combination of these. Also find a basis for the row and column spaces of
the matrices.
1 2 0
1 0 0
3 2 1
0 1 0
(a)
2 1 0 , row echelon form: 0 0 1 . The rank is 3.
0 2 1
0 0 0
1 0 0
1 0 0
4 1 1
0 1 0
(b)
2 1 0 , row echelon form: 0 0 1 . The rank is 3
0 2 0
0 0 0
0 1 0 2 1 2 2
0 1 0 2 0 1 1
0 3 2 12 1 6 8
0 0 1 3 0 1 2
(c)
0 1 1 5 0 2 3 , row echelon form: 0 0 0 0 1 1 1 .
0 2 1 7 0 3 4
0 0 0 0 0 0 0
The rank is 3.
0 1 0 2 0 1 0
0 1 0 2 0 1 0
0 3 2 6 0 5 4
0 0 1 0 0 1 2
(d)
0 1 1 2 0 2 2 , row echelon form: 0 0 0 0 0 0 0 . The
0 2 1 4 0 3 2
0 0 0 0 0 0 0
rank is 2.
0 1 0 2 1 1 2
0 1 0 2 0 1 0
0 3 2 6 1 5 1
0 0 1 0 0 1 0
(e)
0 1 1 2 0 2 1 , row echelon form: 0 0 0 0 1 0 0 . The
0 2 1 4 0 3 1
0 0 0 0 0 0 1
rank is 4.
10. Suppose A is an m × n matrix. Explain why the rank of A is always no larger than
min (m, n) .
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
36
Exercises
It is because you cannot have more than min (m, n) nonzero rows in the row reduced
echelon form. Recall that the number of pivot columns is the same as the number of
nonzero rows from the description of this row reduced echelon form.
11. Suppose A is an m × n matrix in which m ≤ n. Suppose also that the rank of A equals
m. Show that A maps Fn onto Fm . Hint: The vectors e1 , · · · , em occur as columns
in the row reduced echelon form for A.
It follows from the fact that e1 , · · · , em occur as columns in row reduced echelon form
that the dimension of the column space of A is n and so, since this column space is
A (Rn ) , it follows that it equals Fm .
12. Suppose A is an m × n matrix and that m > n. Show there exists b ∈ Fm such that
there is no solution to the equation
Ax = b.
Since m > n the dimension of the column space of A is no more than n and so the
columns of A cannot span Fm .
13. Suppose A is an m × n matrix in which m ≥ n. Suppose also that the rank of A
equals n. Show that A is one to one. Hint: If not, there exists a vector, x 6= 0 such
that Ax = 0, and this implies at least one column of A is a linear combination of the
others. Show this would require the column rank to be less than n.
The hint gives it away. The matrix has at most n independent columns. If one is a
combination of the others, then its rank is less than n.
14. Explain why an n × n matrix A is both one to one and onto if and only if its rank is
n.
If A is one to one, then its columns are linearly independent, hence a basis and so it
is also onto. Since the columns are independent, its rank is n. If its rank is n this just
says that the columns are independent.
15. Suppose A is an m × n matrix and {w1 , · · · , wk } is a linearly independent set of
vectors in A (Fn ) ⊆ Fm . Suppose also that Azi − wi . Show that {z1 , · · · , zk } is also
linearly independent.
P
P
If i ci zi = 0, apply A to both sides to obtain i ci wi = 0. By assumption, each
ci = 0.
16. Let A, B be m × n matrices. Show that rank (A + B) ≤ rank (A) + rank (B).
This is obvious from the following.
rank (A + B) = dim (span (a1 + b1 , · · · , an + bn ))
≤
≤
=
dim (span (a1 , · · · , an , b1 , · · · , bn ))
dim (span (b1 , · · · , bn )) + dim (span (a1 , · · · , an ))
rank (A) + rank (B)
17. Suppose A is an m × n matrix, m ≥ n and the columns of A are independent. Suppose also that {z1 , · · · , zk } is a linearly independent set of vectors in Fn . Show that
{Az1 , · · · , Azk } is linearly independent.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
37
Exercises
Since columns are independent, if Ax = 0, then x = 0. Therefore, the matrix is one
to one. If
n
X
cj Azj = 0,
j=1
then A
P
n
j=1 cj zj
= 0 and so
Pn
j=1 cj zj
= 0 therefore each cj = 0.
18. Suppose A is an m × n matrix and B is an n × p matrix. Show that
dim (ker (AB)) ≤ dim (ker (A)) + dim (ker (B)) .
Hint: Consider the subspace, B (Fp ) ∩ ker (A) and suppose a basis for this subspace
is {w1 , · · · , wk } . Now suppose {u1 , · · · , ur } is a basis for ker (B) . Let {z1 , · · · , zk }
be such that Bzi = wi and argue that
ker (AB) ⊆ span (u1 , · · · , ur , z1 , · · · , zk ) .
Here is how you do this. Suppose ABx = 0. Then Bx ∈ ker (A) ∩ B (Fp ) and so
Pk
Bx = i=1 ai Bzi showing that
x−
k
X
ai zi ∈ ker (B) .
i=1
Let {w1 , · · · , wk } be a basis for B (Fp ) ∩ ker (A) and suppose {u1 , · · · , ur } is a basis
for ker (B). Let Bzi = wi . Now suppose x ∈ ker (AB). Then Bx ∈ ker (A) ∩ B (Fp )
and so
k
k
X
X
Bx =
a i wi =
ai Bzi
i=1
so
x−
i=1
k
X
i=1
Hence
x−
k
X
ai zi ∈ ker (B)
ai zi =
i=1
This shows that
r
X
b j uj
j=1
dim (ker (AB)) ≤ k + r ≤ dim (ker (A)) + dim (ker (B))
Note that a little more can be said. {u1 , · · · , ur , z1 , · · · , zk } is independent. Here is
why. Suppose
k
X
X
a i ui +
bj zj = 0
i
j=1
Then do B to both sides and conclude that
X
bj Bzj = 0
j
which forces each bj = 0. Then the independence of the ui implies each ai = 0. It
follows that in fact,
dim (ker (AB)) = k + r ≤ dim (ker (A)) + dim (ker (B))
and the remaining inequality is an equality if and only if B (Fp ) ⊇ ker (A).
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
38
Exercises
19. Let m < n and let A be an m × n matrix. Show that A is not one to one.
There are more columns than rows and at most m can be pivot columns so it follows
at least one column is a linear combination of the others hence A is not one too one.
20. Let A be an m × n real matrix and let b ∈ Rm . Show there exists a solution, x to the
system
AT Ax = AT b
Next show that if x, x1 are two solutions, then Ax = Ax1 . Hint: First show that
T
AT A = AT A. Next show if x ∈ ker AT A , then Ax = 0. Finally apply the Fredholm alternative. Show AT b ∈ ker(AT A)⊥ . This will give existence of a solution.
T
T
AT A = AT AT
= AT A. Also if AT Ax = 0, then
AT Ax, x = |Ax|2 = 0
⊥
and so Ax = 0. Is AT b ∈ ker AT A ? Say AT Ax = 0. Then Ax = 0. Then
AT b, x = (b,Ax) = (b, 0) = 0
Hence AT b ∈ ker(AT A)⊥ and so there exists a solution to the above system.
21. Show that in the context of Problem 20 that if x is the solution there, then |b − Ax| ≤
|b − Ay| for every y. Thus Ax is the point of A (Rn ) which is closest to b of every
point in A (Rn ). This is a solution to the least squares problem.
|b−Ay|
2
=
=
=
=
|b−Ax+Ax−Ay|
2
2
2
2
2
2
2
|b−Ax| + |Ax − Ay| + 2 (b−Ax,A (x − y))
|b−Ax| + |Ax − Ay| + 2 AT b−AT Ax, (x − y)
|b−Ax| + |Ax − Ay|
and so, Ax is closest to b out of all vectors Ay.
1
0
0 1
T
4
22. Here is a point in R : (1, 2, 3, 4) . Find the point in span
2 , 3
3
2
is closest to the given point.
This just says to find the least squares solution to the
1 0
1
0 1 x
2
=
2 3 y
3
3 2
4
1
0
2
3
T
0
1
0
1
3 2
2
3
0
1
0
1
x
=
2
3 y
2
3
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
T
0
1
2
1
3 3
2
4
which
equations
7KH6D\ORU)RXQGDWLRQ
39
Exercises
14 12
12 14
x
y
=
19
19
, Solution is:
1
19
0 + 19
2
26
26
3
19
26
19
26
. Then the desired point is
0
1
=
3
2
19
26
19
26
95
26
95
26
T
23. ↑Here is a point in R4 : (1, 2, 3, 4) . Find the point on the plane described by x + 2y −
4z + 4w = 0 which is closest to the given point.
−2y + 4z − 4w
y
where y, z, w are arbitary. Thus it is
The plane is of the form
z
w
the span of the vectors
−2
4
−4
1 0 0
0 , 1 , 0
0
0
1
and so you are looking for a least squares solution to the system
−2 4 −4
1
a
1 0 0
b = 2 and to find this, you use the method of least
0 1 0
3
c
0 0 1
4
squares.
T
T
−2 4 −4
−2 4 −4
−2 4 −4
1
a
1 0 0 1 0 0
1 0 0 2
b =
0 1 0 0 1 0
0 1 0 3
c
0 0 1
0 0 1
0 0 1
4
56
5
−8
8
a
0
37
−8 17 −16 b = 7 , Solution is: 147 The closest point is
37
112
8 −16 17
c
0
37
then
28
−2
4
−4
37
147 0 112 0 56
56 1
37
37 0 + 37 1 + 37 0 = 147
37
112
0
0
1
37
28
Check:
1
2 −4 4
37
56
37
147
37
112
37
=0
24. Suppose A, B are two invertible n × n matrices. Show there exists a sequence of row
operations which when done to A yield B. Hint: Recall that every invertible matrix
is a product of elementary matrices.
Both A, B have row reduced echelon form equal to I. Therefore, they are row equivalent.
E1 · · · Ep A = I, F1 · · · Fr B = I
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
40
Exercises
and so
E1 · · · Ep A = F1 · · · Fr B
Now multiply on both sides by the inverses of the Fi to obtain a product of elementary
matrices which multiplied by A give B.
25. If A is invertible and n × n and B is n × p, show that AB has the same null space as
B and also the same rank as B.
From the above problem, AB and B have exactly the same row reduced echelon form.
Thus the result follows.
26. Here are two matrices in row reduced echelon form
1
1 0 1
A = 0 1 1 , B = 0
0
0 0 0
0 0
1 1
0 0
Does there exist a sequence of row operations which when done to A will yield B?
Explain.
No. The row reduced echelon form is unique.
27. Is it true that an upper triagular matrix has rank equal to the number of nonzero
entries down the main diagonal?
1 0 2 0
0 1 1 7
No.
0 0 0 1
0 0 0 0
28. Let {v1 , · · · , vn−1 } be vectors in Fn . Describe a systematic way to obtain a vector vn
which is perpendicular to each of these vectors. Hint: You might consider something
like this
e1
e2
···
en
v11
v12
···
v1n
det
..
..
..
.
.
.
v(n−1)1
where vij is the j
th
v(n−1)2
···
v(n−1)n
entry of the vector vi . This is a lot like the cross product.
This is a good hint. You formally expand along the top row. It will work. If you
replace ei with vki so that the top row becomes.
vk1 vk2 · · · vkn
then the determinant will be equal to 0 because it has two equal rows. However, this
determinant is also what you get when you take the dot product of the above formal
determinant having the ei on the top row with vk .
29. Let A be an m × n matrix. Then ker (A) is a subspace of Fn . Is it true that every
subspace of Fn is the kernel or null space of some matrix? Prove or disprove.
Let M be a subspace of Fn . If it equals {0} , consider the matrix I. Otherwise, it has
a basis {m1 , · · · , mk } . Consider the matrix
m1 · · · mk 0
where 0 is either not there in case k = n or has n − k columns.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
41
Exercises
30. Let A be an n×n matrix and let P ij be the permutation matrix which switches the ith
and j th rows of the identity. Show that P ij AP ij produces a matrix which is similar
to A which switches the ith and j th entries on the main diagonal.
This is easy to see when you consider that P ij is its own inverse and that P ij multiplied
on the right switches the ith and j th columns. Thus you switch the columns and then
you switch the rows. This has the effect of switching Aii and Ajj . For example,
a b c d
1 0 0 0
a d c b
1 0 0 0
0 0 0 1 e f z h 0 0 0 1 n g h t
0 0 1 0 j k l m 0 0 1 0 = j m l k
0 1 0 0
n t h g
0 1 0 0
e h z f
More formally, the iith entry of P ij AP ij is
X ij
ij
Pis Asp Ppi
= Pijij Ajj Pjiij = Aij
s,p
31. Recall the procedure for finding the inverse of a matrix on Page 48. It was shown that
the procedure, when it works, finds the inverse of the matrix. Show that whenever
the matrix has an inverse, the procedure works.
If A has an inverse, then it is one to one. Hence the columns are independent. Therefore, they are each pivot columns. Therefore, the row reduced echelon form of A is I.
This is what was needed for the procedure to work.
F.21
Exercises
5.8
1.
2.
3.
4.
5.
1 2 0
1 0 0
1 2
Find a LU factorization of 2 1 3 . = 2 1 0 0 −3
1 2 3
1 0 1
0 0
1 2 3 2
1
0
0
1
1
0 0
Find a LU factorization of 1 3 2 1 . = 1
5 0 1 3
5 −10 1
0
1 2 1
1 0 0
1 0
Find a P LU factorization of 1 2 2 . = 0 0 1 2 1
2 1 1
0 1 0
1 0
1 2 1 2 1
Find a P LU factorization of 2 4 2 4 1 .
1 2 1 3 2
1 0 0
1 0 0
1 2 1 2 1
= 0 1 0 2 1 0 0 0 0 0 −1
0 0 1
1 0 1
0 0 0 1 1
1 2 1
1 2 2
Find a P LU factorization of
2 4 1 .
3 2 1
1 0 0 0
1 0 0 0
1 2
1
0 0 0 1 3 1 0 0 0 −4 −2
=
0 0 1 0 2 0 1 0 0 0 −1
0 1 0 0
1 0 −1 1
0 0
0
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
0
3
3
3
2
−1 −1
−24 −17
0
1 2
1
0 0 −3 −1
1
0 0
1
2
1
0
7KH6D\ORU)RXQGDWLRQ
42
Exercises
6. Is there only one LU factorization for a given matrix? Hint: Consider the equation
0 1
1 0
0 1
=
.
0 1
1 1
0 0
Apparently, LU factorization
0 1
1 0
0
=
0 1
1 1
0
0 1
1 0
0
=
0 1
0 1
0
7. Here is a matrix and
1
A= 1
0
is not unique.
1
0
1
1
an LU factorization
2 5 0
1
1 4 9 = 1
1 2 5
0
of it.
0 0
1
1 0 0
−1 1
0
Use this factorization to solve the system of equations
1
Ax = 2
3
2
5
0
−1 −1 9
0
1 14
1 0 0
a
1
1
1 1 0 b = 2 , Solution is: 1
0 −1 1
c
3
4
24tˆ4 − 9
x
1 2
5
0
1
y
23tˆ4 − 5
1 , Solution is:
0 −1 −1 9
z =
4 − 14tˆ4
0 0
1 14
4
w
tˆ4
8. Find a QR factorization for the matrix
1 2 1
3 −2 1
1 0 2
1 2 1
3 −2 1
1 0 2
=
√
1
11 √11
3
11 √11
1
11 11
√
11
0
0
9. Find a QR factorization for the matrix
1 2 1
3 0 1
1 0 2
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
√ √
, tˆ4 ∈ R
√
13
2 √11 − 61 √2
66 √
5
− 66√ 2√ 11 − 61√ 2
1
2
33 2 11
3 2
√
√
4
6
−√
11
11 √11
11
√
√
6
2
2 11
11 2 11
11 √
0
·
2
0
1
1
7KH6D\ORU)RXQGDWLRQ
43
Exercises
1 2 1 0
3 0 1 1
1 0 2 1
√ √
1√
1
10 √
11
0 √
11 √11
11 √
√
3
1
3
= 11
√11 − 110
√10√11 −310√ 2√ 5
1
1
11 − 110 10 11 10 2 5
√
√
√
√ 11
2
6
4
11
11
11
11
11
11
11
√
√
√
√
√
√
2
1
2
0
11 10 11
22 √10√ 11 − 55 √ 10
√ 11
1
1
0
0
2
5
2
5
2
5
10. If you had a QR factorization, A = QR, describe how you could use it to solve the
equation Ax = b. Note, this is not how people usually solve systems of equations.
You could first solve Qy = b to obtain y = QT b. Then you would solve Rx = y which
would be easy because R is upper triangular.
11. If Q is an orthogonal matrix, show the columns are an orthonormal set. That is show
that for
Q = q1 · · · qn
it follows that qi · qj = δ ij . Also show that any orthonormal set of vectors is linearly
independent.
Pn
Say i=1 ci qi = 0. Then take the inner product of both sides with qk to obtain
X
ci δ ij = 0
i
Therefore, cj = 0 and so the vectors are independent.
12. Show you can’t expect uniqueness for QR factorizations. Consider
0 0 0
0 0 1
0 0 1
and verify this equals
and also
0
√
1 2
2√
1
2 2
1
0
0 0
√
1
0
2 0 0
2 √
0 0
0 − 12 2
1 0
0 1
0 0
0
0
0 0
1
0
0 0
0 1 .
0 1
Using Definition 5.7.4, can it be concluded that if
follow there is only one QR factorization?
√
0
1
0
0 0
0 0
2
√
√
1
1 2 0
0 0 0 = 0 0
2
2√
2 √
1
1
0 0
0 0 0
2 2 0 −2 2
1 0 0
0 0 0
0 0 0
0 1 0 0 0 1 = 0 0 1
0 0 1
0 0 1
0 0 1
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
√
2
0
0
A is an invertible matrix it will
0
1
1
7KH6D\ORU)RXQGDWLRQ
44
Exercises
Now if A−1 exists and A = QR = Q0 R0 where R, R0 have positive entries down
−1
the main diagonal, then you get R (R0 )
= QT Q0 ≡ U, where U is an orthogonal
matrix. Thus you have an upper triangular matrix which has positive entries down
the diagonal equal to an orthogonal matrix. Obviously, this requires that the upper
triangular matrix is the identity. Hence R = R0 and Q = Q0 . Why is this so obvious?
Consider the following case of a 3×3 which is orthogonal.
a b c
0 d e
0 0 f
Then this matrix
a b
0 d
0 0
a 0
b d
c e
times its transpose is the identity.
2
c
a 0 0
a + b2 + c2 ce + bd cf
e
b d 0
ce + bd
d2 + e2 f e
=
f
c e f
cf
fe
f2
2
0
a b c
a
ab
ac
0 0 d e = ab b2 + d2
de + bc
f
0 0 f
ac de + bc c2 + f 2 + e2
Thus both of these on the right end are the identity and so b = c = 0, a = 1 since
the diagonal entries are all positive. Then also d = 1 and c, e = 0 and f = 1. Now it
follows that all the off diagonal entries must equal 0 since otherwise the magnitude of
a column would not equal 1.
13. Suppose {a1 , · · · , an } are linearly independent vectors in Rn and let
A = a1 · · · an
Form a QR factorization for A.
a1
···
an
=
q1
Show that for each k ≤ n,
···
qn
r11
0
..
.
r12
r22
···
···
..
.
0
0
···
r1n
r2n
rnn
span (a1 , · · · , ak ) = span (q1 , · · · , qk )
Prove that every subspace of Rn has an orthonormal basis. The procedure just described is similar to the Gram Schmidt procedure which will be presented later.
From the way we multiply matrices,
ak ∈ span (q1 , · · · , qk )
Hence span (a1 , · · · , ak ) ⊆ span (q1 , · · · , qk ). In addition to this, each rii > 0 because
the ai are a linearly independent set and if this were not so, then the triangular matrix
on the right would fail to be invertible which would be a contradiction since you could
express it as the product of two invertible matrices. Multiplying on the right by its
inverse, you can get a similar expression for the qi and see that span (q1 , · · · , qk ) ⊆
span (a1 , · · · , ak ) . Thus
span (a1 , · · · , ak ) = span (q1 , · · · , qk )
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
45
Exercises
If you have a subspace, let {a1 , · · · , ak } be a basis and then extend to a basis of
Rn , {a1 , · · · , an }. Then form the above QR factorization and you will have that
{q1 , · · · , qk } is an orthonormal basis for the subspace.
14. Suppose Qn Rn converges to an orthogonal matrix Q where Qn is orthogonal and Rn
is upper triangular having all positive entries on the diagonal. Show that then Qn
converges to Q and Rn converges to the identity.
Let
Q = (q1 , · · · , qn ) , Qk = qk1 , · · · , qkn
k
where the q are the columns. Also denote by rij
the ij th entry of Rk . Thus
It follows
Qk Rk = qk1 , · · · , qkn
k
r11
..
∗
.
0
k
rnn
k k
r11
q1 → q1
and so
k k
k
r11
= r11
q1 → 1
Therefore,
qk1 → q1 .
Next consider the second column.
k k
k k
r12
q1 + r22
q2 → q2
Taking the inner product of both sides with qk1 it follows
k
lim r12
= lim q2 · qk1 = (q2 · q1 ) = 0.
k→∞
k→∞
Therefore,
k k
lim r22
q2 = q2
k→∞
k
k
and since r22
> 0, it follows, as in the first part that r22
→ 1. Hence
lim qk2 = q2 .
k→∞
Continuing this way, it follows
k
lim rij
=0
k→∞
for all i 6= j and
k
lim rjj
= 1, lim qkj = qj .
k→∞
k→∞
Thus Rk → I and Qk → Q.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
46
Exercises
F.22
Exercises
6.6
1. Maximize and minimize z = x1 − 2x2 + x3 subject to the constraints x1 + x2 + x3 ≤
10, x1 + x2 + x3 ≥ 2, and x1 + 2x2 + x3 ≤ 7 if possible. All variables are nonnegative.
The constraints lead to the augmented
1 1 1
1 1 1
1 2 1
The obvious solution is not feasible.
1 1
0 0
0 1
matrix
1
0
0
0 0 10
−1 0 2
0 1 7
Do a row operation.
1 1 0 0 10
0 1 1 0 8
0 −1 0 1 −3
An obvious solution is still not feasible. Do another operation couple of row operations.
1 1 1 0 −1 0 2
0 0 0 1 1 0 8
0 1 0 0 1 1 5
At this point, you can spot an
tableau.
1
0
0
−1
obvious feasible solution. Now assemble the simplex
1 1 0 −1 0 0 2
0 0 1 1 0 0 8
1 0 0 1 1 0 5
2 −1 0 0 0 1 0
Now preserve the simple columns.
1 1
0 0
0 1
0 2
1
0
0
0
0
1
0
0
−1
1
1
−1
0
0
1
0
0
0
0
1
2
8
5
0
First lets work on minimizing this. There is a +2. The ratios are then 5, 2 so the pivot
is the 1 on the top of the second column. The next tableau is
1 1 1 0 −1 0 0 2
0 0 0 1 1 0 0 8
−1 0 −1 0 2 1 0 3
−2 0 −2 0 1 0 1 −4
There is a 1 on the bottom. The ratios of interest for that column are 3/2, 8, and so
the pivot is the 2 in that column. Then the next tableau is
1
1
1
7
1
0 0
0
2
2
2
2
1
13
1
0
1 0 − 12 0
2
2
2
−1 0 −1 0 2 1 0
3
− 32 0 − 23 0 0 − 12 1 − 11
2
Now you stop because there are no more positive numbers to the left of 1 on the
bottom row. The minimum is −11/2 and it occurs when x1 = x3 = x6 = 0 and
x2 = 7/2, x4 = 13/2, x6 = −11/2.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
47
Exercises
Next consider maximization. The simplex tableau was
1 1 1 0 −1 0 0
0 0 0 1 1 0 0
0 1 0 0 1 1 0
−1 2 −1 0 0 0 1
2
8
5
0
This time you work on getting rid of the negative entries. Consider the −1 in the first
column. There is only one ratio to consider so 1 is the pivot.
1 1 1 0 −1 0 0 2
0 0 0 1 1 0 0 8
0 1 0 0 1 1 0 5
0 3 0 0 −1 0 1 2
There remains a −1. The ratios are 5 and 8 so the next pivot is the 1 in the third row
and column 5.
1 2 1 0 0 1 0 7
0 −1 0 1 0 −1 0 3
0 1 0 0 1 1 0 5
0 4 0 0 0 1 1 7
Then no more negatives remain so the maximum is 7 and it occurs when x1 = 7, x2 =
0, x3 = 0, x4 = 3, x5 = 5, x6 = 0.
2. Maximize and minimize the following if possible. All variables are nonnegative.
(a) z = x1 − 2x2 subject to the constraints x1 + x2 + x3 ≤ 10, x1 + x2 + x3 ≥ 1, and
x1 + 2x2 + x3 ≤ 7
an augmented matrix for the constraints is
1 1 1 1 0 0 10
1 1 1 0 −1 0 1
1 2 1 0 0 1 7
The obvious solution is not feasible. Do some row operations.
0 0 0 1 1 0 9
1 1 1 0 −1 0 1
−1 0 −1 0 2 1 5
Now the obvious solution is feasible.
0 0 0
1 1 1
−1 0 −1
−1 2 0
First preserve the simple columns.
0 0 0
1 1 1
−1 0 −1
−3 0 −2
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
Then include the objective function.
1 1 0 0 9
0 −1 0 0 1
0 2 1 0 5
0 0 0 1 0
1
0
0
0
1
−1
2
2
0
0
1
0
0
0
0
1
9
1
5
−2
7KH6D\ORU)RXQGDWLRQ
48
Exercises
Lets try to maximize first. Begin with the first
Use it.
0 0 0 1 1 0
1 1 1 0 −1 0
0 1 0 0 1 1
0 3 1 0 −1 0
There is still a -1 on the bottom row to the
so the new pivot is the 1 on the third row.
0 −1 0 1 0
1 2 1 0 0
0 1 0 0 1
0 4 1 0 0
column. The only pivot is the 1.
0 9
0 1
0 6
1 1
left of the 1. The ratios are 9 and 6
−1
1
1
1
Then the maximum is 7 when x1 = 7 and x1 , x3
Next consider the minimum.
0 0 0 1 1 0
1 1 1 0 −1 0
−1 0 −1 0 2 1
−3 0 −2 0 2 0
There is a positive 2 in the bottom row left of 1.
2.
1
0 21 1 0 − 12
2
1
1 1 1 0 0
2
2
2
−1 0 −1 0 2 1
−2 0 −1 0 0 −1
0
0
0
1
= 0.
3
7
6
7
9
1
5
−2
0
0
0
1
The pivot in that column is the
0 13
2
0 72
0 5
1 −7
The minimum is −7 and it happens when x1 = 0, x2 = 7/2, x3 = 0.
(b) z = x1 − 2x2 − 3x3 subject to the constraints x1 + x2 + x3 ≤ 8, x1 + x2 + 3x3 ≥ 1,
and x1 + x2 + x3 ≤ 7
This time, lets use artificial variables to find an initial simplex tableau. Thus you
add in an artificial variable and then do a minimization procedure.
1
1
1
0
1
1
1
0
1
3
1
0
1
0
0
0
First preserve the seventh column
1 1 1
1 1 3
1 1 1
1 1 3
Now use the third column.
2 2
3
1
2
3
0
3
1
2
3
0
0
3
0
0
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
0
−1
0
0
0 0
0 1
1 0
0 −1
0
0
0
1
8
1
7
0
as a simple column by a row operation.
1 0 0 0 0 8
0 −1 0 1 0 1
0 0 1 0 0 7
0 −1 0 0 1 1
1 13 0
0 −1 0
0 13 1
0 0 0
− 31
1
− 31
−1
0
0
0
1
23
3
1
0
20
3
7KH6D\ORU)RXQGDWLRQ
49
Exercises
It follows that a basic solution is feasible if
2 2
0 1 13 0
3
3
1 1 3 0 −1 0
2
2
0 0 13 1
3
3
Now assemble the simplex tableau
2
2
0
3
3
1 1 3
2
2
0
3
3
−1 2 3
1 13 0
0 −1 0
0 13 1
0 0 0
23
3
1
20
3
0
0
0
1
23
3
1
0
20
3
Preserve the simple columns by doing row operations.
2
2
0 1 13 0 0 23
3
3
3
1 1 3 0 −1 0 0 1
2
2
0 0 13 1 0 20
3
3
3
−2 1 0 0 1 0 1 −1
Lets do minimization first.
0
1
0
−3
Work with the second column.
0 −2 1 1 0 0 7
1 3 0 −1 0 0 1
0 −2 0 1 1 0 6
0 −3 0 2 0 1 −2
Recall how you have to pick the pivot correctly.
in the bottom row left of the 1. Work with that
0 0 0 1 0 −1
1 1 1 0 0 1
0 0 −2 0 1 1
−3 0 1 0 0 −2
There is still a positive number
column.
0
1
0
7
0
6
1 −14
There is still a positive number to the left of 1 on the bottom row.
0
0 0 1 0 −1 0
1
1
1 1 0 0 1 0
7
2
2 0 0 1 3 0 20
−4 −1 0 0 0 −3 1 −21
It follows that the minimum is −21 and it occurs when x1 = x2 = 0, x3 = 7.
Next consider the maximum. The simplex tableau was
2
2
0 1 13 0 0 23
3
3
3
1 1 3 0 −1 0 0 1
2
2
0 0 13 1 0 20
3
3
3
−2 1 0 0 1 0 1 −1
Use the first column.
0
1
0
0
0
1
0
3
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
−2
3
−2
6
1
0
0
0
1
−1
1
−1
0
0
1
0
0
0
0
1
7
1
6
1
7KH6D\ORU)RXQGDWLRQ
50
Exercises
There is still a negative on the bottom row to the left of 1.
0 0 0 1 0 −1 0 1
1 1 1 0 0 1 0 7
0 0 −2 0 1 1 0 6
0 3 4 0 0 1 1 7
There are no more negatives on the bottom row left of 1 so stop. The maximum
is 7 and it occurs when x1 = 7, x2 = 0, x3 = 0.
(c) z = 2x1 + x2 subject to the constraints x1 − x2 + x3 ≤ 10, x1 + x2 + x3 ≥ 1, and
x1 + 2x2 + x3 ≤ 7.
The augmented matrix for the constraints is
1 −1 1 1 0 0 10
1 1 1 0 −1 0 1
1 2 1 0 0 1 7
The basic solution is not feasible because of that −1. Lets do a row operation to
change this. I used the 1 in the second column as a pivot and zeroed out what
was above and below it. Now it seems that the basic solution is feasible.
2 0 2 1 −1 0 11
1 1 1 0 −1 0 1
−1 0 −1 0 2 1 5
Assemble the simplex tableau.
2
0
2 1 −1
1
1
1 0 −1
−1 0 −1 0 2
−2 −1 0 0 0
0
0
1
0
0
0
0
1
11
1
5
0
Then do a row operation to
2
1
−1
−1
preserve the simple columns.
0 2 1 −1 0 0 11
1 1 0 −1 0 0 1
0 −1 0 2 1 0 5
0 1 0 −1 0 1 1
It follows that the minimum
Now lets maximize.
2
1
−1
−1
is 0 and it occurs when x1 = x2 = 0, x3 = 1.
Lets do minimization first. Work with the third column because there is a positive
entry on the bottom.
0 −2 0 1 1 0 0 9
1
1 1 0 −1 0 0 1
0
1 0 0 1 1 0 6
−2 −1 0 0 0 0 1 0
0
1
0
0
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
2
1
−1
1
1 −1 0
0 −1 0
0 2 1
0 −1 0
0 11
0 1
0 5
1 1
7KH6D\ORU)RXQGDWLRQ
51
Exercises
Lets begin with the first column.
0 −2
1 1
0 1
0 1
There is still a −2 to the left of
0 −3
1 2
0 1
0 3
0
1
0
2
1
0
0
0
1
−1
1
−2
0
0
1
0
0
0
0
1
1 in the bottom row.
0
1
0
2
1
0
0
0
9
1
6
2
3
7
6
14
0 −1 0
0 1 0
1 1 0
0 2 1
There are no negatives left so the maximum is 14 and it happens when x1 =
7, x2 = x3 = 0.
(d) z = x1 + 2x2 subject to the constraints x1 − x2 + x3 ≤ 10, x1 + x2 + x3 ≥ 1, and
x1 + 2x2 + x3 ≤ 7.
The augmented matrix for the constraints is
1 −1 1 1 0 0 10
1 1 1 0 −1 0 1
1 2 1 0 0 1 7
Of course the obvious or basic solution is not feasible. Do a row operation involving a pivot in the second row to try and fix this.
2 0 2 1 −1 0 11
1 1 1 0 −1 0 1
−1 0 −1 0 2 1 5
Now all is well. Begin to assemble the
2
0
2
1
1
1
−1 0 −1
−1 −2 0
simplex tableau.
1 −1
0 −1
0 2
0 0
0
0
1
0
0
0
0
1
11
1
5
0
Do a row operation to preserve the simple columns.
2 0 2 1 −1 0 0 11
1 1 1 0 −1 0 0 1
−1 0 −1 0 2 1 0 5
1 0 2 0 −2 0 1 2
Next lets maximize. There is only one negative
3
0 23 1 0 12
2
1 1 1 0 0 1
2
2
2
−1 0 −1 0 2 1
0 0 1 0 0 1
number in the bottom left of 1.
0 27
2
0 72
0 5
1 7
Thus the maximum is 7 and it happens when x2 = 7/2, x3 = x1 = 0.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
52
Exercises
Next lets find the minimum.
2
1
−1
1
Start with the column which
0
1
0
−1
0
1
0
0
2
1
−1
2
1 −1 0
0 −1 0
0 2 1
0 −2 0
0 11
0 1
0 5
1 2
1
0
0
0
0
0
0
1
has a 2.
−2
1
1
−2
0
1
0
0
1
−1
1
0
0
0
1
0
9
1
6
0
There are no more positive numbers so the minimum is 0 when x1 = x2 = 0, x3 =
1.
3. Consider contradictory constraints, x1 + x2 ≥ 12 and x1 + 2x2 ≤ 5. You know these
two contradict but show they contradict using the simplex algorithm.
You can do this by using artificial variables, x5 .
1 1 −1 0 1
1 2 0 1 0
0 0 0 0 −1
Thus
0 12
0 5
1 0
Do a row operation to preserve the simple columns.
1 1 −1 0 1 0 12
1 2 0 1 0 0 5
1 1 −1 0 0 1 12
Next start with the 1 in the first column.
0 −1 −1 −1 1
1 2
0
1 0
0 −1 −1 −1 0
0 7
0 5
1 7
Thus the minimum value of z = x5 is 7 but, for there to be a feasible solution, you
would need to have this minimum value be 0.
4. Find a solution to the following inequalities for x, y ≥ 0 if it is possible to do so. If it
is not possible, prove it is not possible.
(a)
6x + 3y ≥ 4
8x + 4y ≤ 5
Use an artificial variable. Let x1 = x, x2
artificial
variable x5 . Then minimize
x5 as
6 3 −1 0 1 0 4
8 4 0 1 0 0 5
0 0 0 0 −1 1 0
Keep the simple columns.
6 3 −1 0
8 4 0 1
6 3 −1 0
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
= y and slack variables x3 , x4 with
described earlier.
1
0
0
0 4
0 5
1 4
7KH6D\ORU)RXQGDWLRQ
53
Exercises
Now proceed to minimize.
0 0
8 4
0 0
−1 − 43
0
1
−1 − 43
It appears that the minimum value
these inequalities with x1 , x2 ≥ 0.
6x1 + 4x3 ≤ 11
(b) 5x1 + 4x2 + 4x3 ≥ 8
6x1 + 6x2 + 5x3 ≤ 11
The augmented matrix is
6 0 4
5 4 4
6 6 5
1
0
0
1
4
0
0
1
5
1
4
of x5 is 1/4 and so there is no solution to
1
0
0
0 0 11
−1 0 8
0 1 11
It is not clear whether there is a solution which has all variables nonnegative.
However, if you do a row operation using 5 in the first column as a pivot, you get
0 − 24
− 45 1 65 0 75
5
5
4
4 0 −1 0 8
6
1
0
0 65 1 75
5
5
and so a solution is x1 = 8/5, x2 = x3 = 0.
6x1 + 4x3 ≤ 11
(c) 5x1 + 4x2 + 4x3 ≥ 9
6x1 + 6x2 + 5x3 ≤ 9
The augmented matrix is
6 0 4 1 0 0 11
5 4 4 0 −1 0 9
6 6 5 0 0 1 9
Lets include an artificial variable
6 0 4
5 4 4
6 6 5
0 0 0
Preserving the simple columns,
6 0 4
5 4 4
6 6 5
5 4 4
Use the first column.
0 −6 −1
0 −1 − 1
6
6 6
5
0 −1 − 16
and seek to minimize x7 .
1 0 0 0 0 11
0 −1 0 1 0 9
0 0 1 0 0 9
0 0 0 −1 1 0
0
1
0
0
0
0
0
1
11
9
9
9
0 −1
−1 − 65
0
1
−1 − 65
0
1
0
0
0
0
0
1
1 0 0
0 −1 0
0 0 1
0 −1 0
1
0
0
0
2
3
2
9
3
2
It appears that the minimum value for x7 is 3/2 and so there will be no solution
to these inequalities for which all the variables are nonnegative.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
54
Exercises
(d)
x1 − x2 + x3 ≤ 2
x1 + 2x2 ≥ 4
3x1 + 2x3 ≤ 7
The augmented matrix is
1
1
3
−1 1 1
2 0 0
0 2 0
0 0
−1 0
0 1
2
4
7
Lets add in an artificial variable and set things up to minimize this artificial
variable.
1 −1 1 1 0 0 0 0 2
1 2 0 0 −1 0 1 0 4
3 0 2 0 0 1 0 0 7
0 0 0 0 0 0 −1 1 0
Then
1 −1 1
1 2 0
3 0 2
1 2 0
Work with second column.
3
2
1
3
0
0
2
0
0
1
0
2
0
1 0 0
0 −1 0
0 0 1
0 −1 0
1
0
0
0
− 21
−1
0
0
0
0
1
0
0
0
0
1
2
4
7
4
0
1 0
0 0
−1 1
4
4
7
0
0
1
0
0
1
2
It appears the minimum value of x7 is 0 and so this will mean there is a solution
when x2 = 2, x3 = 0, x1 = 0.
(e)
5x1 − 2x2 + 4x3 ≤ 1
6x1 − 3x2 + 5x3 ≥ 2
5x1 − 2x2 + 4x3 ≤ 5
The augmented matrix is
5
6
5
lets introduce an artificial
5
6
5
0
−2 4 1
−3 5 0
−2 4 0
0 0
−1 0
0 1
1
2
5
variable x7 and then minimize x7 .
−2 4 1 0 0 0 0 1
−3 5 0 −1 0 1 0 2
−2 4 0 0 1 0 0 5
0 0 0 0 0 −1 1 0
Then, preserving the simple columns,
5 −2 4 1 0 0
6 −3 5 0 −1 0
5 −2 4 0 0 1
6 −3 5 0 −1 0
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
0
1
0
0
0
0
0
1
1
2
5
2
7KH6D\ORU)RXQGDWLRQ
55
Exercises
work with the first column.
5 −2
0 −3
5
0 0
0 − 53
4
1
5
0
1
5
There is still a positive entry.
5
−2
−1 −1
2
4
0
0
− 14 − 21
1
− 56
−1
− 56
4
0
0
0
0
−1
0
−1
1
− 45
−1
− 45
0
0
1
0
0
−1
0
−1
0
0
1
0
0
1
0
0
0
1
0
0
0
0
0
1
0
0
0
1
1
4
5
4
4
5
1
3
4
4
3
4
It appears that there is no solution to this system of inequalities because the
minimum value of x7 is not 0.
5. Minimize z = x1 + x2 subject to x1 + x2 ≥ 2, x1 + 3x2 ≤ 20, x1 + x2 ≤ 18. Change
to a maximization problem and solve as follows: Let yi = M − xi . Formulate in terms
of y1 , y2 .
You could find the maximum of 2M − x1 − x2 for the given constraints and this would
happen when x1 + x2 is as small as possible. Thus you would maximize y1 + y2 subject
to the constraints
M − y1 + M − y2
M − y1 + 3 (M − y2 )
M − y1 + M − y2
≥ 2
≤ 20
≤ 18
To simplify, this would be
2M − 2
≥ y1 + y2
4M − 20 ≤ y1 + 3y2
2M − 18 ≤ y1 + y2
You could simply regard M as large enough that yi ≥ 0 and use the techniques just
developed. The augmented matrix for the constraints is then
1 1 1 0
0
2M − 2
1 3 0 −1 0 4M − 20
1 1 0 0 −1 2M − 18
Here M is large. Use the 3 as a pivot to zero out above and below it.
2
3
1
2
3
0
3
0
Then it is still the case that the
row and pick the 2/3 as a pivot.
0 0
0 3
2
0
3
1 31
0
0 −1 0
0 31 −1
+ 14
3
4M − 20
2
34
3M − 3
2
3M
basic solution is not feasible. Lets use the bottom
1 0
0 − 32
1
0
3
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
1
3
2
−1
16
3M − 3
2
34
3M − 3
7KH6D\ORU)RXQGDWLRQ
56
Exercises
Now it appears that the basic solution is feasible provided M
the simplex tableau.
0
0 1 0
1 0
16
3
3
0
3
0
−
0
3M
−3
2
2
2
1
2
34
0
0
−1
0
M
−
3
3
3
3
−1 −1 0 0
0 1
0
Do row operations to preserve
0 0
0 3
2
0
3
0 0
the simple columns.
1 0
0 − 23
0 31
0 0
1
0
3
0
2
−1 0
−1 1
is large. Then assemble
16
3M − 3
2
34
3M − 3
2M − 18
There is a negative number to the left of the 1 on the bottom row and we want to
maximize so work with this column. Assume M is very large. Then the pivot should
be the top entry in this column.
0 0 1
0 1 0
16
0 3 − 3 − 3 0 0 3M − 27
2
2
2
1
0 1
0 0 32 M + 14
3
3
3
0 0 1
0 0 1 2M − 2
It follows that the maximum of y1 +y2 is 2M −2 and it happens when y1 = M +7, y2 =
M − 9, y3 = 0. Thus the minimum of x1 + x2 is
M − y1 + M − y2 = M − (M + 7) + M − (M − 9) = 2
F.23
Exercises
7.3
1. If A is the matrix of a linear transformation which rotates all vectors in R2 through
30◦ , explain why A cannot have any real eigenvalues.
Because the vectors which result are not parallel to the vector you begin with.
2. If A is an n × n matrix and c is a nonzero constant, compare the eigenvalues of A and
cA.
λ → cλ
3. If A is an invertible n × n matrix, compare the eigenvalues of A and A−1 . More
generally, for m an arbitrary integer, compare the eigenvalues of A and Am .
λ → λ−1 and λ → λm .
4. Let A, B be invertible n × n matrices which commute. That is, AB = BA. Suppose
x is an eigenvector of B. Show that then Ax must also be an eigenvector for B.
Say Bx = λx. Then
B (Ax) = A (Bx) = Aλx = λ (Ax)
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
57
Exercises
5. Suppose A is an n × n matrix and it satisfies Am = A for some m a positive integer
larger than 1. Show that if λ is an eigenvalue of A then |λ| equals either 0 or 1.
Let x be the eigenvector. Then Am x = λm x,Am x = Ax = λx and so
λm = λ
Hence if λ 6= 0, then
λm−1 = 1
and so |λ| = 1.
6. Show that if Ax = λx and Ay = λy, then whenever a, b are scalars,
A (ax + by) = λ (ax + by) .
Does this imply that ax + by is an eigenvector? Explain.
The formula is obvious from properties of matrix multiplications. However, this vector
might not be an eigenvector because it might equal 0 and
Eigenvectors are never 0
7. Find the eigenvalues and eigenvectors of
−1
−1
−1
the matrix
−1 7
0 4 .
−1 5
Determine
whether
defective.
the matrix is
−1 −1 7
3
2
−1 0 4 , eigenvectors:
1
1
↔ 1,
↔ 2. This is a defective matrix.
−1 −1 5
1
1
8. Find the eigenvalues and eigenvectors of the matrix
−3 −7 19
−2 −1 8 .
−2 −3 10
Determine
whether
the matrix is defective.
−3 −7 19
3
1
2
−2 −1 8 , eigenvectors: 1 ↔ 1, 2 ↔ 2, 1 ↔ 3
−2 −3 10
1
1
1
This matrix has distinct eigenvalues so it is not defective.
9. Find the eigenvalues and eigenvectors of the matrix
−7 −12 30
−3 −7 15 .
−3 −6 14
Determine
whetherthe matrix is defective.
−7 −12 30
5
−2
2
−3 −7 15 , eigenvectors: 1 , 0 ↔ −1, 1 ↔ 2
−3 −6 14
0
1
1
This matrix is not defective because, even though λ = 1 is a repeated eigenvalue, it
has a 2 dimensional eigenspace.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
58
Exercises
10. Find the eigenvalues and eigenvectors of
7
8
−2
the matrix
−2 0
−1 0 .
4 6
Determine whether the matrix is
defective.
7 −2 0
− 21
0
8 −1 0 , eigenvectors: −1 ↔ 3, 0 ↔ 6
−2 4 6
1
1
This matrix is defective.
11. Find the eigenvalues and eigenvectors of the
3 −2
0 5
0 2
matrix
−1
1 .
4
Determine whether the matrix is
defective.
0
3 −2 −1
1
−1
0 5
1 , eigenvectors: 0 , − 21 ↔ 3, 1 ↔ 6
0
0 2
4
1
1
This matrix is not defective.
12. Find the eigenvalues and eigenvectors of the matrix
6 8 −23
4 5 −16
3 4 −12
Determine whether the matrix is defective.
13. Find the eigenvalues and eigenvectors of the matrix
5 2 −5
12 3 −10 .
12 4 −11
Determine
5 2
12 3
12 4
whether the matrix is
defective.
5
−5
− 13
6
−10 , eigenvectors: 1 , 0 ↔ −1
−11
0
1
This matrix is defective. In this case, there is only one eigenvalue, −1 of multiplicity
3 but the dimension of the eigenspace is only 2.
14. Find the eigenvalues and eigenvectors of the matrix
20 9 −18
6 5 −6 .
30 14 −27
Determine whether the matrix is defective.
20 9 −18
6
5 −6 , eigenvectors:
30 14 −27
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
59
Exercises
1
9
2
13
↔ −1, 1 ↔ 2, 3 ↔ −3
13
1
1
1
3
4
1
4
Not defective.
15. Find the eigenvalues and eigenvectors of the matrix
1
26 −17
4
−4
4 .
−9 −18
9
Determine whether the matrix is defective.
16. Find the eigenvalues and eigenvectors of the
3 −1
11 3
8
0
matrix
−2
−9 .
−6
Determine whether the matrix is defective.
3
3 −1 −2
4
11 3 −9 , eigenvectors: 1 ↔ 0 This one is defective.
4
8
0 −6
1
17. Find the eigenvalues and eigenvectors of the matrix
−2
1 2
−11 −2 9 .
−8
0 7
Determine whether the matrix is defective.
3
−2
1 2
4
−11 −2 9 , eigenvectors: 1 ↔ 1
4
−8
0 7
1
This is defective.
18. Find the eigenvalues and eigenvectors of the matrix
2 1 −1
2 3 −2 .
2 2 −1
Determine whether the matrix is defective.
1
2 1 −1
1
−1
2
2 3 −2 , eigenvectors: 1 , 0 ↔ 1, 1 ↔ 2
1
2 2 −1
0
1
This is non defective.
19. Find the complex eigenvalues and eigenvectors of the matrix
4 −2 −2
0 2 −2 .
2 0
2
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
60
Exercises
4 −2 −2
0 2 −2 , eigenvectors:
2 0
2
1
−i
i
−1 ↔ 4, −i ↔ 2 − 2i, i ↔ 2 + 2i
1
1
1
20. Find the eigenvalues and eigenvectors of the matrix
9
6 −3
0
6
0 .
−3 −6 9
Determine whether the matrix is defective.
1
9
6 −3
−1
−2
0
6
0 , eigenvectors: 1 , 0 ↔ 6, 0 ↔ 12
1
1
0
−3 −6 9
This is nondefective.
4 −2 −2
21. Find the complex eigenvalues and eigenvectors of the matrix 0 2 −2 . De2 0
2
termine whether the matrix is defective.
4 −2 −2
0 2 −2 , eigenvectors:
2 0
2
1
−i
i
−1 ↔ 4, −i ↔ 2 − 2i, i ↔ 2 + 2i
1
1
1
−4 2
0
22. Find the complex eigenvalues and eigenvectors of the matrix 2 −4 0 .
−2 2 −2
Determine whether the matrix is defective.
−4 2
0
2 −4 0 , eigenvectors:
−2 2 −2
1
1
0
0 , 1 ↔ −2, −1 ↔ −6
1
0
1
This is not defective.
1
1 −6
23. Find the complex eigenvalues and eigenvectors of the matrix 7 −5 −6 .
−1 7
2
Determine whether the matrix is defective.
1
1 −6
7 −5 −6 , eigenvectors:
−1 7
2
1
−i
i
−1 ↔ −6, −i ↔ 2 − 6i, i ↔ 2 + 6i
1
1
1
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
61
Exercises
This is not defective.
4 2 0
24. Find the complex eigenvalues and eigenvectors of the matrix −2 4 0 . Deter−2 2 6
mine whether the matrix is defective.
4 2 0
−2 4 0 , eigenvectors:
−2 2 6
1
1
0
0 ↔ 6, −i ↔ 4 − 2i, i ↔ 4 + 2i
1
1
1
This is not defective.
25. Here is a matrix.
1
0
0
0
0
0
c
2
a 0
1 b
0 2
0 0
Find values of a, b, c for which the matrix is defective and values of a, b, c for which it
is nondefective.
First consider the eigenvalue λ = 1
0 a
0 0
0 0
0 0
0
b
1
0
1
b
0
0
0
c
1
0
0
0
0
Then you have ax2 = 0, bx3 = 0. If neither a nor b = 0 then λ = 1 would be a defective
eigenvalue and the matrix would be defective. If a = 0, then the dimension of the
eigenspace is clearly 2 and so the matrix would be nondefective. If b = 0 but a 6= 0,
then you would have a defective matrix because the eigenspace would have dimension
less than 2. If c 6= 0, then the matrix is defective. If c = 0 and a = 0, then it is non
defective. Basically, if a, c 6= 0, then the matrix is defective.
26. Here is a matrix.
a
0
0
0
1
c
where a, b, c are numbers. Show this is sometimes defective depending on the choice
of a, b, c. What is an easy case which will ensure it is not defective?
An easy case which will ensure that it is not defective is for a, b, c to be distinct. If
you have any repeats, then this will be defective.
27. Suppose A is an n × n matrix consisting entirely of real entries but a + ib is a complex
eigenvalue having the eigenvector, x + iy. Here x and y are real vectors. Show that
then a − ib is also an eigenvalue with the eigenvector, x − iy. Hint: You should
remember that the conjugate of a product of complex numbers equals the product of
the conjugates. Here a + ib is a complex number whose conjugate equals a − ib.
A (x + iy) = (a + ib) (x + iy) . Now just take complex conjugates of both sides.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
62
Exercises
28. Recall an n × n matrix is said to be symmetric if it has all real entries and if A = AT .
Show the eigenvalues of a real symmetric matrix are real and for each eigenvalue, it
has a real eigenvector.
Let λ be an eigenvalue which corresponds to x 6= 0. Then λxT = xT AT . Then
¯
λxT x
¯ = xT AT x
¯ = xT A¯
x = xT x
¯λ
¯ = λ. We have A¯
which shows that λ
x = λ¯
x and Ax = λx so
A (¯
x + x) = λ (¯
x + x)
Hence it has a real eigenvector.
29. Recall an n × n matrix is said to be skew symmetric if it has all real entries and if
A = −AT . Show that any nonzero eigenvalues must be of the form ib where i2 = −1.
In words, the eigenvalues are either 0 or pure imaginary.
Let A be skew symmetric. Then if x is an eigenvector for λ,
¯
λxT x
¯ = xT AT x
¯ = −xT A¯
x = −xT x
¯λ
¯ Thus a + ib = − (a − ib) and so a = 0.
and so λ = −λ.
30. Is it possible for a nonzero matrix to have only 0 as an eigenvalue?
−12 −16
Sure. Try this one.
9
12
31. Show that the eigenvalues and eigenvectors of a real matrix occur in conjugate pairs.
This follows from the observation that if Ax = λx, then Ax = λx
32. Suppose A is an n × n matrix having all real eigenvalues which are distinct. Show
there exists S such that S −1 AS = D, a diagonal matrix. If
λ1
0
..
D=
.
0
define eD by
and define
eD ≡
λn
eλ1
0
..
0
.
eλn
eA ≡ SeD S −1 .
Next show that if A is as just described, so is tA where t is a real number and the
eigenvalues of At are tλk . If you differentiate a matrix of functions entry by entry so
that for the ij th entry of A0 (t) you get a0ij (t) where aij (t) is the ij th entry of A (t) ,
show
d At e
= AeAt
dt
Next show det eAt 6= 0. This is called the matrix exponential. Note I have only
defined it for the case where the eigenvalues of A are real, but the same procedure will
work even for complex eigenvalues. All you have to do is to define what is meant by
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
63
Exercises
ea+ib .
It is clear that the eigenvalues of tA are tλ where λ is an eigenvalue of A. Thus eAt is
of the form
tλ
e 1
0
−1
..
S
S
.
0
etλn
And so
d At e
=
dt
S
= S
0
λ1
..
0
..
0
.
0
λ1
.
λn
= AetA
0
..
S
=
λ1 etλ1
−1
S
.
λn etλn
tλ1
0
e
λn
0
−1
S S
..
0
.
etλn
etλ1
0
..
.
0
etλn
33. Find the principle directions determined by the matrix
eigenvalues are 31 , 1, and 12 listed according to multiplicity.
1
1
−2
1
−1 , 1 , 1 , 1 , 1 , 1
2
2
3
1
0
1
7
12
− 41
1
6
−1
S
−1
S
− 14
7
12
− 16
1
6
− 61
2
3
. The
34. Find the principle directions determined by the matrix
5
− 31 − 31
3
7
1
−1
The eigenvalues are 1, 2, and 1. What is the physical interpreta3
6
6
1
1
7
−3
6
6
tion of the repeated eigenvalue?
There are apparently two directions in which there is no stretching.
1 1
−2
2
2
1 , 0 ↔ 1, 1 ↔ 2
0
1
1
35. Find oscillatory solutions to the system of differential equations, x00 = Ax where A =
−3 −1 −1
−1 −2 0 The eigenvalues are −1, −4, and −2.
−1 0 −2
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
64
Exercises
The eigenvalues are given and so there are oscillatory solutions of the form
−1
1 (a cos (t) + b sin (t)) ,
1
0
√ √ −1 c sin
2t + d cos
2t
1
2
1 (e cos (2t) + f sin (2t))
1
where a, b, c, d, e, f are scalars.
36. Let A and B be n × n matrices and let the columns of B be
b1 , · · · , bn
and the rows of A are
aT1 , · · · , aTn .
Show the columns of AB are
Ab1 · · · Abn
and the rows of AB are
aT1 B · · · aTn B.
The ith entry of the j th column of AB is aTi bj . Therefore, these
columns are as
indicated. Also the ith row of AB will be aTi b1 · · · aTi bn . Thus this row is
just aTi B.
37. Let M be an n × n matrix. Then define the adjoint of M , denoted by M ∗ to be the
transpose of the conjugate of M. For example,
∗ 2
i
2 1−i
=
.
1+i 3
−i
3
A matrix M, is self adjoint if M ∗ = M. Show the eigenvalues of a self adjoint matrix
are all real.
∗
First note that (AB) = B ∗ A∗ . Say M x = λx, x 6= 0. Then
2
∗
∗
λ |x| = λx∗ x = (λx) x = (M x) x = x∗ M ∗ x
= x∗ M x = x∗ λx = λ |x|
2
Hence λ = λ.
38. Let M be an n × n matrix and suppose x1 , · · · , xn are n eigenvectors which form a
linearly independent set. Form the matrix S by making the columns these vectors.
Show that S −1 exists and that S −1 M S is a diagonal matrix (one having zeros everywhere except on the main diagonal) having the eigenvalues of M on the main diagonal.
When this can be done the matrix is said to be diagonalizable.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
65
Exercises
Since the vectors are linearly independent, the matrix S has an inverse. Denoting this
inverse by
T
w1
..
−1
S = .
wnT
it follows by definition that
wiT xj = δ ij .
Therefore,
w1T
S −1 M S = S −1 (M x1 , · · · , M xn ) = ... (λ1 x1 , · · · , λn xn )
wnT
λ1
..
=
0
.
0
λn
39. Show that a n × n matrix M is diagonalizable if and only if Fn has a basis of eigenvectors. Hint: The first part is done in Problem 38. It only remains to show that if the
matrix can be diagonalized by some matrix S giving D = S −1 M S for D a diagonal
matrix, then it has a basis of eigenvectors. Try using the columns of the matrix S.
The formula says that
M S = SD
Letting xk denote the k th column of S, it follows from the way we multiply matrices
that
M xk = λk xk
where λk is the k th diagonal entry on D.
40. Let
and let
1
3
A=
0
B=
2
4
1
0 1
1 1
2 1
2
0
3
Multiply AB verifying the block multiplication formula. Here A11 =
2
, A21 = 0 1 and A22 = (3) .
0
1 2 2
0 1
6 5
AB = 3 4 0 1 1 = 4 7
0 1 3
2 1
7 4
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
1
3
2
4
, A12 =
7KH6D\ORU)RXQGDWLRQ
66
Exercises
Now consider doing it by block multiplication. This leads to
1 2
0 1
2
2
1
+
3 4
0
1 1
0 1
0 1
+3 2 1
1 1
6 5
6 5
4 7 = 4 7
=
7 4
7 4
41. Suppose A, B are n × n matrices and λ is a nonzero eigenvalue of AB. Show that then
it is also an eigenvalue of BA. Hint: Use the definition of what it means for λ to be
an eigenvalue. That is,
ABx = λx
where x 6= 0. Maybe you should multiply both sides by B.
B (ABx) = λBx Hence (BA) (Bx) = λ (Bx) . Is Bx = 0? No, it can’t be since λ 6= 0.
If Bx = 0, then the left side of the given equation would be 0 and the right wouldn’t
be.
42. Using the above problem show that if A, B are n × n matrices, it is not possible that
AB − BA = aI for any a 6= 0. Hint: First show that if A is a matrix, then the
eigenvalues of A − aI are λ − a where λ is an eigenvalue of A.
If it is true that AB − BA = aI for a 6= 0, Then you get AB − aI = BA. If
{λ1 , · · · , λn } are the eigenvalues of AB, then the eigenvalues of AB − aI = BA are
{λ1 − a, · · · , λn − a} which cannot be the same set, contrary to the above problem.
43. Consider the following matrix.
0
1
C =
0
···
0
..
.
0
..
.
−a0
−a1
..
.
−an−1
1
Show det (λI − C) = a0 + λa1 + · · · an−1 λn−1 + λn . This matrix is called a companion
matrix for the given polynomial.
λI − C is of the form
λ
−1
0
0
λ
..
.
···
..
.
−1
λ + an−1
Expand along the bottom row.
λ
−1
(λ + an−1 ) det
0
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
a0
a1
..
.
0
λ
..
.
..
.
−1
λ
−1 λ
7KH6D\ORU)RXQGDWLRQ
67
Exercises
λ
0
−1 λ
−1
+1 det
0
That last determinant is of the
λ
−1
det
0
···
0
λ
..
.
..
.
−1
a0
a1
a2
..
.
an−2
form
0
λ
−1
0
λ
..
.
···
a0
a1
a2
..
.
..
.
−1
λ + (an−2 − λ)
By induction, this is of the form
λn−1 + (an−2 − λ) λn−2 + · · · + a1 λ + a0
and so, the sum of the two is of the form
λn−1 (λ + an−1 ) + λn−1 + (an−2 − λ) λn−2 + · · · + a1 λ + a0
= λn + an−1 λn−1 + an−2 λn−2 + · · · + a1 λ + a0
It is routine to verify that when n = 2 this determinant gives the right polynomial.
44. A discreet dynamical system is of the form
x (k + 1) = Ax (k) , x (0) = x0
where A is an n × n matrix and x (k) is a vector in Rn . Show first that
x (k) = Ak x0
for all k ≥ 1. If A is nondefective so that it has a basis of eigenvectors, {v1 , · · · , vn }
where
Avj = λj vj
you can write the initial condition x0 in a unique way as a linear combination of these
eigenvectors. Thus
n
X
x0 =
aj vj
j=1
Now explain why
x (k) =
n
X
aj Ak vj =
j=1
n
X
aj λkj vj
j=1
which gives a formula for x (k) , the solution of the dynamical system.
The first formula is obvious from induction. Thus
Ak x0 =
n
X
j=1
aj Ak vj =
n
X
aj λkj vj
j=1
because if Av = λv, then if Ak x = λk x, do A to both sides. Thus Ak+1 x = λk+1 x.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
68
Exercises
45. Suppose A is an n × n matrix and let v be an eigenvector such that Av = λv. Also
suppose the characteristic polynomial of A is
det (λI − A) = λn + an−1 λn−1 + · · · + a1 λ + a0
Explain why
An + an−1 An−1 + · · · + a1 A + a0 I v = 0
If A is nondefective, give a very easy proof of the Cayley Hamilton theorem based on
this. Recall this theorem says A satisfies its characteristic equation,
An + an−1 An−1 + · · · + a1 A + a0 I = 0.
Let v be any vector in a basis of eigenvectors of A. Since A is nondefective, such a
basis exists. Then
An + an−1 An−1 + · · · + a1 A + a0 I v = λn + an−1 λn−1 + · · · + a1 λ + a0 v = 0
because λ is by definition a solution to the characteristic polynomial. Since the above
holds for vectors in a basis, it holds for any v and so
An + an−1 An−1 + · · · + a1 A + a0 I = 0
46. Suppose an n × n nondefective matrix A has only 1 and −1 as eigenvalues. Find A12 .
P
Let x be some vector. Let x = j aj vj where vj is an eigenvector. Then
A12 x =
n
X
aj λ12
j vj =
j=1
n
X
aj vj = x
j=1
It follows that A12 = I.
47. Suppose the characteristic polynomial of an n × n matrix A is 1 − λn . Find Amn where
m is an integer. Hint: Note first that A is nondefective. Why?
The eigenvalues are distinct because they are the nth roots of 1. Hence from the above
formula, if x is a given vector with
x=
n
X
aj vj
j=1
then
nm
A
nm
x=A
n
X
aj vj =
j=1
so Anm = I.
n
X
j=1
nm
aj A
vj =
n
X
aj vj = x
j=1
48. Sometimes sequences come in terms of a recursion formula. An example is the Fibonacci sequence.
x0 = 1 = x1 , xn+1 = xn + xn−1
Show this can be considered as a discreet dynamical system as follows.
xn+1
1 1
xn
x1
1
=
,
=
xn
1 0
xn−1
x0
1
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
69
Exercises
Now use the technique of Problem 44 to find a formula for xn .
What are the eigenvalues and eigenvectors of this matrix?
1 1
, eigenvectors:
1 0
1 1 √ 1 √
1
1√
1
1 1√
2 − 2 5
2 5+ 2
5,
↔
5+
↔ −
1
1
2 2
2
2
Now also
1
1√
−
5
2 10
1
2
−
1
2
1
√ 5
+
Therefore, the solution is of the form
1√
1
5+
10
2
xn+1
xn
1
2
√
5+
1
1
2
=
1
1
=
n 1 1 √ 1√
1 1√
1
2 − 2 5
−
5
−
5
1
2 10
2 2
n 1 √
1
√
√
1
1
1
1
2 5+ 2
+
5+
5+
1
10
2
2
2
In particular,
xn =
n n
1√
1 1√
1√
1
1√
1
1
−
5
−
5 +
5+
5+
2 10
2 2
10
2
2
2
49. Let A be an n × n matrix having characteristic polynomial
det (λI − A) = λn + an−1 λn−1 + · · · + a1 λ + a0
n
Show that a0 = (−1) det (A).
The characteristic polynomial equals det (λI − A) . To get the constant term, you plug
in λ = 0 and obtain det (−A) = (−1)n det (A).
F.24
Exercises
7.10
1. Explain why it is typically impossible to compute the upper triangular matrix whose
existence is guaranteed by Schur’s theorem.
To get it, you must be able to get the eigenvalues and this is typically not possible.
2. Now recall the QR factorization of Theorem 5.7.5 on Page 133. The QR algorithm
is a technique which does compute the upper triangular matrix in Schur’s theorem.
There is much more to the QR algorithm than will be presented here. In fact, what
I am about to show you is not the way it is done in practice. One first obtains what
is called a Hessenburg matrix for which the algorithm will work better. However,
the idea is as follows. Start with A an n × n matrix having real eigenvalues. Form
A = QR where Q is orthogonal and R is upper triangular. (Right triangular.) This
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
70
Exercises
can be done using the technique of Theorem 5.7.5 using Householder matrices. Next
take A1 ≡ RQ. Show that A = QA1 QT . In other words these two matrices, A, A1 are
similar. Explain why they have the same eigenvalues. Continue by letting A1 play the
role of A. Thus the algorithm is of the form An = QRn and An+1 = Rn+1 Q. Explain
why A = Qn An QTn for some Qn orthogonal. Thus An is a sequence of matrices each
similar to A. The remarkable thing is that often these matrices converge to an upper
triangular matrix T and A = QT QT for some orthogonal matrix, the limit of the Qn
where the limit means the entries converge. Then the process computes the upper
triangular Schur form of the matrix A. Thus the eigenvalues of A appear on the
diagonal of T. You will see approximately what these are as the process continues.
Suppose you have found An . Then An = Qn Rn and then
An+1 ≡ Rn Qn = QTn An Qn = QTn QTn−1 An−1 Qn−1 Qn · · ·
Now the product of orthogonal matrices is orthogonal and so this shows by induction
that each An is orthogonally similar to A.
3. Try the QR algorithm on
−1 −2
6
6
which has eigenvalues 3 and 2. I suggest you use a computer algebra system to do the
computations.
√
√
√ √
1
6
−1 −2
− 37
37 − 37
37
37 38
37
√
√
√37
=
6
6
1
6
6
− 37 37
0
37 37
37 37
√ √
√ 191
√
1
38
6
− 37√ 37 − 37 √37
− 260
37 37 √37
37
37
A1 =
=
36
6
6
6
1
− 37
− 37
37
0
37
37 37
37 37
191
− 260
37
37
=
36
6
− 37
37
√ √
√ √
191 √ √
1 √ √
36
1348
1021
37
1021
37 777 √37√1021 − 37 777√ 37
37 37 1021 − 37 777
√
√
√
36
191
6
0
37 777 37 1021
37 777 37 1021
1021 37 1021
3959
7952
− 1021
1021
A2 =
216
1146
1021
1021
3959
7952
−
1021
1021
=
0.211 56 1146
1021
0.998 51
−5. 447 9 × 10−2
3. 883 3 −7. 715 7
5. 447 9 × 10−2
0.998 51
0
1. 545 1
3. 883 3 −7. 715 7
0.998 51
−5. 447 9 × 10−2
A3 =
=
0
1. 545 1
5. 447 9 × 10−2
0.998 51
3. 457 2
−7. 915 8
8. 417 6 × 10−2 1. 542 8
You can keep on going. You see that it appears to be converging to an upper triangular
matrix having 3, 1 down the diagonal.
4. Now try the QR algorithm on
0
2
−1
0
Show that the algorithm cannot converge for this example. Hint: Try a few iterations
of the algorithm.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
Exercises
71
−1
0 −1
2 0
=
0
1 0
0 1
2 0
0 −1
0 −2
A1 =
=
0 1
1 0
1 0
0 −2
0 −1
1 0
=
1 0
1 0
0 2
1 0
0 −1
0 −1
A2 =
=
. Now it is back to where you started.
0 2
1 0
2 0
0 −2
0 −1
and
Thus the algorithm merely bounces between the two matrices
1 0
2 0
and so it can’t possibly converge.
0 −1
0 −2
5. Show the two matrices A ≡
and B ≡
are similar; that is
4 0
2 0
there exists a matrix S such that A = S −1 BS but there is no orthogonal matrix
Q such that QT BQ = A. Show the QR algorithm does converge for the matrix B
although it fails to do so for A.
0
2
Both matrices have distinct eigenvalues ±2 and so they are both similar to a diagonal
matrix having these eigenvalues on the diagonal. Therefore, the matrices are indeed
similar. Lets try the QR algorithm on the second.
0 −2
0 −1
2 0
=
2 0
1 0
0 2
2 0
0 −1
0 −2
A1 =
=
and so the algorithm keeps on returning
0 2
1 0
2 0
the same matrix in the case of B. Now consider the matrix A.
0 −1
0 −1
4 0
=
4 0
1 0
0 1
4 0
0 −1
0 −4
A1 =
=
0 1
1 0
1 0
0 −4
0 −1
1 0
=
1 0
1 0
0 4
1 0
0 −1
0 −1
A2 =
=
0 4
1 0
4 0
0 −1
0 −4
In this case, the algorithm bounces between
and
.
4 0
1 0
6. Let F be an m × n matrix. Show that F ∗ F has all real eigenvalues and furthermore,
they are all nonnegative.
The eigenvalues are real because the matrix F ∗ F is Hermitian. If λ is one of them
with eigenvector x,
λ (x, x) = (F ∗ F x, x) = (F x,F x) ≥ 0
7. If A is a real n × n matrix and λ is a complex eigenvalue of A having eigenvector
z + iw, show that w 6= 0.
A (z + iw) = (a + ib) (z + iw)
A (z − iw) = (a − ib) (z − iw)
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
72
Exercises
Now subtract to obtain A (2iw) = 2iaw + 2ibz. Now if w = 0, then you would have
0 =ibz and this requires b = 0 since otherwise, the eigenvector was equal to 0 which
it isn’t. Hence λ was not complex after all.
8. Suppose A = QT DQ where Q is an orthogonal matrix and all the matrices are real.
Also D is a diagonal matrix. Show that A must be symmetric.
AT = QT DT Q = QT DQ = A.
9. Suppose A is an n × n matrix and there exists a unitary matrix U such that
A = U ∗ DU
where D is a diagonal matrix. Explain why A must be normal.
A∗ A = U ∗ D∗ U U ∗ DU = U ∗ D∗ DU
AA∗ = U ∗ DU U ∗ D∗ U = U ∗ DD∗ U. But D∗ D = DD∗ and both are equal to the
diagonal matrix which has the squares of the absolute values of the diagonal entries
of D down the diagonal.
10. If A is Hermitian, show that det (A) must be real.
A = U ∗ DU where D
is a real diagonal matrix. Therefore,
det (A) = det U −1 det (U ) det (D) = det (D) ∈ R.
11. Show that every unitary matrix preserves distance. That is, if U is unitary,
|U x| = |x| .
|U x|2 = (U x, U x) = (U ∗ U x, x) = (x, x) = |x|2 .
12. Show that if a matrix does preserve distances, then it must be unitary.
Let U preserve distances. Then
2
2
|x| + |y| + 2 Re (x, y) =
2
|U (x + y)| = (U x + U y,U x + U y)
2
2
= |U x| + |U y| + 2 Re (U x, U y)
= |x|2 + |y|2 + 2 Re (U x, U y)
Therefore,
Re (U x, U y) − Re (x, y)
∗
Re (U U x − x, y)
= 0
= 0
Let |θ| = 1 and θ (U ∗ U x − x, y) = |(U ∗ U x − x, y)| . Then replace y in the above with
¯θy. Then you get
Re U ∗ U x − x,¯
θy = Re θ (U ∗ U x − x, y) = |(U ∗ U x − x, y)| = 0
and since y is arbitrary,
(U ∗ U x − x, y) = 0
for all y. In particular, this holds for y =U ∗ U x − x and so U ∗ U = I.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
73
Exercises
13. Show that a complex normal matrix A is unitary if and only if its eigenvalues have
magnitude equal to 1.
Since A is normal, it follows that there is a unitary matrix U such that
AU = U D
where D is a diagonal matrix. Then from the above problems, A is unitary if and only
if it preserves distances. Let the columns of U be the orthonormal set {u1 , · · · , un } .
Then we have Auk = λk uk . For A to preserve distances, you must have |λk | = 1.
Conversely, if |λk | = 1 for all k, then {λk uk } is also an orthonormal set and so A
preserves distances and is therefore normal.
14. Suppose A is an n × n matrix which is diagonally dominant. Recall this means
X
|aij | < |aii |
j6=i
show A−1 must exist.
This follows from Gerschgorin’s theorem. The condition implies that 0 is not an
eigenvalue and so the matrix is one to one and hence invertible.
15. Give some disks in the complex plane whose union contains all the eigenvalues of the
matrix
1 + 2i 4 2
0
i 3
5
6 7
B (1 + 2i, 6) , B (i, 3) , B (7, 11)
16. Show a square matrix is invertible if and only if it has no zero eigenvalues.
To say the matrix has no zero eigenvalues is to say that it is one to one because it
maps no nonzero vector to 0.
17. Using Schur’s theorem, show the trace of an n × n matrix equals the sum of the
eigenvalues and the determinant of an n × n matrix is the product of the eigenvalues.
Let A be an n × n matrix. By Schur, there exists a unitary matrix U and an upper
triangular matrix T such that
A = U ∗T U
Then det (A) = det (T ) = the product of the diagonal entries of T. These are
eigenvalues of T and both A and T have the same eigenvalues because they have
same characteristic polynomial. Thus the determinant of A equals the product of
eigenvalues. Since A and T are similar, they have the same trace also. However,
trace of T is just the sum of the eigenvalues of A.
the
the
the
the
18. Using Schur’s theorem, show that if A is any
nP
× n matrix having eigenvalues
P complex
2
n
2
{λi } listed according to multiplicity, then i,j |Aij | ≥ i=1 |λi | . Show that equality
holds if and only if A is normal.
By Schur’s theorem, there is a unitary U and upper triangular T such that U ∗ AU = T.
Then
X
|Aij |2 = trace (AA∗ ) = trace (U T U ∗ U T ∗ U ∗ )
i,j
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
74
Exercises
= trace (U T T ∗U ∗ ) = trace (T T ∗) ≥
n
X
i=1
|λi |
2
Equality holds if and only if there are no nonzero off diagonal terms. This happens if
and only if T is a diagonal matrix. But if this is so, then A must be normal because
U ∗ AU = D, a diagonal matrix and this implies that A is normal.
19. Here is a matrix.
1234
6
5
3
0
−654
9
123
98
123 10, 000 11
56
78
98
400
I know this matrix has an inverse before doing any computations. How do I know?
Gerschgorin’s theorem shows that there are no zero eigenvalues and so the matrix is
invertible.
20. Show the critical points of the following function are
1
(0, −3, 0) , (2, −3, 0) , and 1, −3, −
3
and classify them as local minima, local maxima or saddle points.
f (x, y, z) = − 32 x4 + 6x3 − 6x2 + zx2 − 2zx − 2y 2 − 12y − 18 − 32 z 2 .
3 4
3 2
3
2
2
2
0
= ∇ − 2 x + 6x − 6x2 + zx3 − 2zx− 2y − 12y − 18 − 2 z .
2xz − 2z − 12x + 18x − 6x = 0
After much fuss, you find the above solutions
−4y − 12 = 0
x2 − 2x − 3z = 0
to these systems of equations.
−18x2 + 36x + 2z − 12 0 2x − 2
0
−4
0
Hessian is
2x − 2
0
−3
(0, −3, 0) ,
−12 0
0
−4
−2
0
−2
0
−3
2
0
−3
The eigenvalues are all negative and so this point is a local maximum.
(2, −3, 0)
−12 0
0
−4
2
0
The eigenvalues are all negative so this is a local maximum.
1, −3, − 31
16
0
0
3
0 −4 0
0
0 −3
This is a saddle point.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
75
Exercises
21. Here is a function of three variables.
f (x, y, z) = 13x2 + 2xy + 8xz + 13y 2 + 8yz + 10z 2
change the variables so that in the new variables there are no mixed terms, terms
involving xy, yz etc. Two eigenvalues are 12 and 18.
The function is
T
x
13 1 4
x
f (x, y, z) = y 1 13 4 y
z
4 4 10
z
13 1 4
1 13 4 , eigenvectors:
10
4 1 4
−2
−1
1
− 1 ↔ 6, 1 ↔ 12, 1 ↔ 18
2
1
0
1
Then in terms of the new variables, the quadratic form is 6x02 + 12y 02 + 18z 02. Note
how I only needed to find the eigenvalues. This changes in the next problems.
22. Here is a function of three variables.
f (x, y, z) = 2x2 − 4x + 2 + 9yx − 9y − 3zx + 3z + 5y 2 − 9zy − 7z 2
change the variables so that in the new variables there are no mixed terms, terms
involving xy, yz etc.
The function is f (x, y, z) =
T
x
x
2
9/2 −3/2
x
y 9/2
5
−9/2 y + −4 −9 3 y + 2
z
z
−3/2 −9/2 −7
z
2
9/2 −3/2
9/2
5
−9/2 , eigenvectors:
−7
−3/2 −9/2
5
0
−2
−3 ↔ −1, 1 ↔ − 17 , −3 ↔ 19
2
2
3
1
1
1
√
1√
35
0
− 71 √14
7 √
√
3
1
3
Appropriate orthogonal matrix: Q = − 35
35 10
10 − 14
√
√
√ 14
1
3
1
35
10
14
35
10
14
−1
0
0
−1
0
0
, A = Q 0 − 17 0 QT Thus you need to have
QT AQ = 0 − 17
0
2
2
19
19
0
0
0
0
2
2
the new variables be
√
T
0
1√
35
0
− 71 √14
x
x
7 √
√
− 3 35 1 10 − 3 14 y = y 0
35√
10 √
14√
1
3
1
z
z0
35 35
10 10
14 14
Then in terms of the new variables, the quadratic form is
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
76
Exercises
T
0
−1
0
0
x0
x
y 0 0 − 17 0 y 0 +
2
19
z0
z0
0
0
2
√
1√
0
35
0
− 17 √14
x
7 √
√
3
1
3
y0 + 2
−4 −9 3 − 35
35
10
−
14
10 √
14√
√
1
3
1
z0
35 35
10 10
14 14
√ √
√ √ 0
19 0 2
19
0 2
=−(x0 )2 + 27 5 7x0 − 17
2 7z + 2
2 (y ) + 2 (z ) + 7
23. Here is a function of three variables.
f (x, y, z) = −x2 + 2xy + 2xz − y 2 + 2yz − z 2 + x
change the variables so that in the new variables there are no mixed terms, terms
involving xy, yz etc.
The function is
T
x
−1
y 1
z
1
−1 1
1
1 −1 1
1
1 −1
1
1
x
x
−1 1 y + 1 0 0 y
1 −1
z
z
−1
−1
1
, eigenvectors: 1 ↔ 1, 1 , −1 ↔ −2
0
2
1
√
√
1√
1
1
3 √3 − 2√ 2 − 6 √6
Appropriate orthogonal matrix: Q = 13 √3 21 2 − 16√ 6 .
1
1
0
3 3
3 6
New variables,
x0
y0 =
z0
√
1
3 √3
1
3 √3
1
3 3
√
− 21√ 2
1
2 2
0
√ T
− 61 √6
x
− 61√ 6 y
1
z
3 6
Quadratic form in the new variables:
0 T
0
x
1 0
0
x
y 0 0 −2 0 y 0
z0
0 0 −2
z0
√
√
1√
1
1
x0
3 √3 − 2√ 2 − 6 √6
1
+ 1 0 0 13 √3
− 61√ 6 y 0
2 2
1
1
z0
0
3 3
3 6
√
√
√
=(x0 )2 + 13 3x0 − 2(y 0 )2 − 12 2y 0 − 2(z 0 )2 − 16 6z 0
24. Show the critical points of the function,
f (x, y, z) = −2yx2 − 6yx − 4zx2 − 12zx + y 2 + 2yz.
are points of the form,
(x, y, z) = t, 2t2 + 6t, −t2 − 3t
for t ∈ R and classify them as local minima, local maxima or saddle points.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
77
Exercises
−6y − 12z − 4xy − 8xz
∇ −2yx2 − 6yx − 4zx2 − 12zx + y 2 + 2yz = −2x2 − 6x + 2y + 2z
−4x2 − 12x + 2y
−6y − 12z − 4xy − 8xz = 0
−2x2 − 6x + 2y + 2z = 0 Let x = t. Then y = 2t2 + 6t. Now place these into the
−4x2 − 12x + 2y = 0
middle equation and you then find that −2t2 − 6t + 2 2t2 + 6t + 2z = 0, Solution is:
z = −t2 − 3t. This also works in the top equation.
−6 2t2 + 6t − 12 −t2 − 3t − 4t 2t2 + 6t − 8t −t2 − 3t = 0
−2yx2 − 6yx − 4zx2 − 12zx + y 2 + 2yz, Hessian is
−4y − 8z −4x − 6 −8x − 12
−4x − 6
2
2
−8x − 12
2
0
Plug in the critical points.
0
−4t − 6 −8t − 12
−4t − 6
You need to consider the sign of the eigenvalues.
2
2
−8t − 12
2
0
−λ
−4t − 6 −8t − 12
= 80t2 λ + 240tλ − λ3 + 2λ2 + 184λ. Thus you
2−λ
2
det −4t − 6
−8t − 12
2
−λ
need to be looking at the solutions to
λ3 − 2λ2 − 80t2 λ − 240tλ − 184λ = 0
You get an eigenvalue which equals 0. Then you need to solve
λ2 − 2λ − 184 + 80t2 + 240t = 0
184 + 80t2 + 240t has a minimum value when t = − 23 and at this point, the value of
2
this function is 184 + 80 32 + 240 − 32 = 4. Therefore, from the quadratic formula,
there is a positive eigenvalue and a negative eigenvalue for any value of t. Thus this
function always has a direction in which it appears to have a local min. and a direction
in which it appears to have a local max. for each of the critical points.
25. Show the critical points of the function
1 4
1
x − 4x3 + 8x2 − 3zx2 + 12zx + 2y 2 + 4y + 2 + z 2 .
2
2
are (0, −1, 0) , (4, −1, 0) , and (2, −1, −12) and classify them as local minima, local
maxima or saddle points.
6x2 − 24x − 6z + 16 0 12 − 6x
0
4
0
Hessian:
12 − 6x
0
1
f (x, y, z) =
(0, −1, 0)
16 0
0 4
12 0
12
0
1
This is a saddle point because it has a negative and two positive eigenvalues.
(4, −1, 0)
22 0
0
4
−12 0
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
−12
0
1
7KH6D\ORU)RXQGDWLRQ
78
Exercises
This is also a saddle point because it has a negative and two positive eigenvalues.
(2, −1, −12)
64 0 0
0 4 0
0 0 1
This is a local minimum because all eigenvalues are positive.
26. Let f (x, y) = 3x4 − 24x2 + 48 − yx2 + 4y. Find and classify the critical points using
the second derivative test.
12x3 − 2xy − 48x
4
2
2
∇ 3x − 24x + 48 − yx + 4y =
4 − x2
Critical points: (2, 0) , (−2, 0) .
4
2
2
3x − 24x + 48 − yx + 4y, Hessian is
(2, 0)
36x2 − 2y − 48 −2x
−2x
0
96 −4
−4 0
Clearly it has eigenvalues of different sign because the determinant is negative. Therefore, this is a saddle point. (−2, 0) works out the same way. To have some fun, graph
this function of two variables and see if you can see the critical points are saddle points
from the picture.
27. Let f (x, y) = 3x4 − 5x2 + 2 − y 2 x2 + y 2 . Find and classify the critical points using the
second derivative test.
12x3 − 2xy 2 − 10x
4
2
2 2
2
∇ 3x − 5x + 2 − y x + y =
2y − 2x2 y
√ √ √ √ Critical points: (1, 1) , (−1, 1) , (1, −1) , (−1, −1) , (0, 0) , − 61 5 6, 0 , 16 5 6, 0
36x2 − 2y 2 − 10 −4xy
3x4 − 5x2 + 2 − y 2 x2 + y 2 , Hessian is
−4xy
2 − 2x2
(1, 1)
24 −4
−4 0
Saddle point. The eigenvalues have opposite sign.
(−1, 1) , saddle point also. (1, −1) saddle point (−1, −1) saddle point.
√ √ − 61 5 6, 0
20 0
0 13
√ √ Local minimum. 16 5 6, 0 also a local minimum.
28. Let f (x, y) = 5x4 − 7x2 − 2 − 3y 2 x2 + 11y 2 − 4y 4 . Find and classify the critical points
using the second derivative test.
20x3 − 6xy 2 − 14x
∇ 5x4 − 7x2 − 2 − 3y 2 x2 + 11y 2 − 4y 4 =
−6x2 y − 16y 3 + 22y
√
√
1√ √
1
7 10, 0 , 10
7 10, 0
Critical points (1, 1) , (−1, 1) , (0, 0) , − 10
5x4 − 7x2 − 2 − 3y 2 x2 + 11y 2 − 4y 4 ,
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
79
Exercises
Hessian is
(1, 1)
60x2 − 6y 2 − 14
−12xy
−12xy
−6x2 − 48y 2 + 22
40 −12
−12 −32
The determinant is negative and so this is a saddle point.
(−1, 1)
40 12
12 −32
The determinant is negative and so this is also a saddle point.
(0, 0)
This is a saddle point.
√ √
1
10 7 10, 0
−14 0
0
22
28 0
0 89
5
√ √
1
This is a local minimum. So is − 10 7 10, 0 .
29. Let f (x, y, z) = −2x4 − 3yx2 + 3x2 + 5x2 z + 3y 2 − 6y + 3 − 3zy + 3z + z 2 . Find and
classify the critical points using the second derivative test.
∇ −2x4 − 3yx2 + 3x2 + 5x2 z + 3y 2 − 6y + 3 − 3zy + 3z + z 2
6x − 6xy + 10xz − 8x3
= −3x2 + 6y − 3z − 6
5x2 − 3y + 2z + 3
Critical points: (0, 1, 0)
−24x2 − 6y + 10z + 6 −6x 10x
−6x
6
−3 At the point of interest this is
Hessian is
10x
−3
2
−54 0
0
0
6 −3
0
−3 2
It has two positive eigenvalues and one negative eigenvalue so this is a saddle point.
30. Let f (x, y, z) = 3yx2 − 3x2 − x2 z − y 2 + 2y − 1 + 3zy − 3z − 3z 2 . Find and classify
the critical points using the second derivative test.
∇ 3yx2 − 3x2 − x2 z − y 2 + 2y − 1 + 3zy − 3z − 3z 2
6xy − 6x − 2xz
= 3x2 − 2y + 3z + 2
−x2 + 3y − 6z − 3
Critical points: (0, 1, 0)
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
80
Exercises
6y − 2z − 6
6x
Hessian is
−2x
eigenvalues:
6x −2x
−2
3
3
−6
0
0
0
At the critical point, this is
0
−2
3
√
√
13 − 4, − 13 − 4, 0
0
3
−6
This has two directions in which it appears to have a local maximum. However, the
test fails because of the zero eigenvalue.
31. Let Q be orthogonal. Find the possible values of det (Q) .
±1
32. Let U be unitary. Find the possible values of det (U ) .
U is unitary which means it preserves length. Therefore, if U x = λx, it must be the
case that |λ| = 1. Hence λ = eiθ for some θ. In other words, λ is a point on the unit
circle.
33. If a matrix is nonzero can it have only zero for eigenvalues?
−1 1
Definitely yes. Consider this one.
−1 1
34. A matrix A is called nilpotent if Ak = 0 for some positive integer k. Suppose A is a
nilpotent matrix. Show it has only 0 for an eigenvalue.
Say λ is an eigenvalue. Then for some x 6= 0,
Ax = λx
Then λk is an eigenvalue for Ak and so λk = 0. Hence λ = 0.
35. If A is a nonzero nilpotent matrix, show it must be defective.
If it is not defective, then there exists S such that
S −1 AS = 0
But then A = 0. Hence, if A 6= 0 is nilpotent, then A is defective.
36. Suppose A is a nondefective n × n matrix and its eigenvalues are all either 0 or 1.
Show A2 = A. Could you say anything interesting if the eigenvalues were all either
0,1,or −1? By DeMoivre’s theorem, an nth root of unity is of the form
2kπ
2kπ
cos
+ i sin
n
n
Could you generalize the sort of thing just described to get An = A? Hint: Since A
is nondefective, there exists S such that S −1 AS = D where D is a diagonal matrix.
S −1 AS = D and so S −1 A2 S = D2 = D = S −1 AS and so, in the first case, A2 = A.
If the eigenvaluse are 0,1,−1, then all even powers of A are equal and all odd powers
are equal.
In the last case, you could have the eigenvalues of A equal to cos 2kπ
+ i sin 2kπ
.
n
n
Then
S −1 An S = I
and so An = I.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
81
Exercises
37. This and the following problems will present most of a differential equations course.
To begin with, consider the scalar initial value problem
y 0 = ay, y (t0 ) = y0
When a is real, show the unique solution to this problem is y = y0 ea(t−t0 ) . Next
suppose
y 0 = (a + ib) y, y (t0 ) = y0
(6.30)
where y (t) = u (t) + iv (t) . Show there exists a unique solution and it is
y (t) = y0 ea(t−t0 ) (cos b (t − t0 ) + i sin b (t − t0 ))
≡ e(a+ib)(t−t0 ) y0 .
(6.31)
Next show that for a real or complex there exists a unique solution to the initial value
problem
y 0 = ay + f, y (t0 ) = y0
and it is given by
y (t) = ea(t−t0 ) y0 + eat
Z
t
e−as f (s) ds.
t0
Hint: For the first part write as y 0 − ay = 0 and multiply both sides by e−at . Then
explain why you get
d −at
e y (t) = 0, y (t0 ) = 0.
dt
Now you finish the argument. To show uniqueness in the second part, suppose
y 0 = (a + ib) y, y (0) = 0
and verify this requires y (t) = 0. To do this, note
y 0 = (a − ib) y, y (0) = 0
and that
d
2
|y (t)|
dt
=
y 0 (t) y (t) + y 0 (t) y (t)
=
(a + ib) y (t) y (t) + (a − ib) y (t) y (t)
=
2a |y (t)|2 , |y|2 (t0 ) = 0
2
Thus from the first part |y (t)| = 0e−2at = 0. Finally observe by a simple computation
that 6.30 is solved by 6.31. For the last part, write the equation as
y 0 − ay = f
and multiply both sides by e−at and then integrate from t0 to t using the initial
condition.
38. Now consider A an n × n matrix. By Schur’s theorem there exists unitary Q such that
Q−1 AQ = T
where T is upper triangular. Now consider the first order initial value problem
x0 = Ax, x (t0 ) = x0 .
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
82
Exercises
Show there exists a unique solution to this first order system. Hint: Let y = Q−1 x
and so the system becomes
y0 = T y, y (t0 ) = Q−1 x0
(6.32)
T
Now letting y = (y1 , · · · , yn ) , the bottom equation becomes
yn0 = tnn yn , yn (t0 ) = Q−1 x0 n .
Then use the solution you get in this to get the solution to the initial value problem
which occurs one level up, namely
0
yn−1
= t(n−1)(n−1) yn−1 + t(n−1)n yn , yn−1 (t0 ) = Q−1 x0 n−1
Continue doing this to obtain a unique solution to 6.32.
39. Now suppose Φ (t) is an n × n matrix of the form
Φ (t) =
x1 (t) · · ·
xn (t)
where
(6.33)
x0k (t) = Axk (t) .
Explain why
Φ0 (t) = AΦ (t)
if and only if Φ (t) is given in the form of 6.33. Also explain why if c ∈ Fn ,
y (t) ≡ Φ (t) c
solves the equation
y0 (t) = Ay (t) .
This is because y0 (t) = Φ0 (t) c = AΦ (t) c = Ay (t) .
40. In the above problem, consider the question whether all solutions to
x0 = Ax
(6.34)
are obtained in the form Φ (t) c for some choice of c ∈ Fn . In other words, is the
general solution to this equation Φ (t) c for c ∈ Fn ? Prove the following theorem using
linear algebra.
Theorem F.24.1 Suppose Φ (t) is an n × n matrix which satisfies
Φ0 (t) = AΦ (t) .
−1
Then the general solution to 6.34 is Φ (t) c if and only if Φ (t)
exists for some t.
−1
−1
Furthermore, if Φ0 (t) = AΦ (t) , then either Φ (t) exists for all t or Φ (t) never
exists for any t.
(det (Φ (t)) is called the Wronskian and this theorem is sometimes called the Wronskian
alternative.)
Hint: Suppose first the general solution is of the form Φ (t) c where c is an arbitrary
−1
constant vector in Fn . You need to verify Φ (t) exists for some t. In fact, show
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
83
Exercises
−1
−1
Φ (t) exists for every t. Suppose then that Φ (t0 )
there exists c ∈ Fn such that there is no solution x to
does not exist. Explain why
c = Φ (t0 ) x
By the existence part of Problem 38 there exists a solution to
x0 = Ax, x (t0 ) = c
but this cannot be in the form Φ (t) c. Thus for every t, Φ (t)−1 exists. Next suppose
−1
for some t0 , Φ (t0 ) exists. Let z0 = Az and choose c such that
z (t0 ) = Φ (t0 ) c
Then both z (t) , Φ (t) c solve
x0 = Ax, x (t0 ) = z (t0 )
Apply uniqueness to conclude z = Φ (t) c. Finally, consider that Φ (t) c for c ∈ Fn
−1
either is the general solution or it is not the general solution. If it is, then Φ (t)
−1
exists for all t. If it is not, then Φ (t) cannot exist for any t from what was just
shown.
−1
41. Let Φ0 (t) = AΦ (t) . Then Φ (t) is called a fundamental matrix if Φ (t)
t. Show there exists a unique solution to the equation
x0 = Ax + f , x (t0 ) = x0
exists for all
(6.35)
and it is given by the formula
−1
x (t) = Φ (t) Φ (t0 )
x0 + Φ (t)
Z
t
−1
Φ (s)
f (s) ds
t0
Now these few problems have done virtually everything of significance in an entire undergraduate differential equations course, illustrating the superiority of linear algebra.
The above formula is called the variation of constants formula.
Hint: Uniquenss is easy. If x1 , x2 are two solutions then let u (t) = x1 (t) − x2 (t) and
argue u0 = Au, u (t0 ) = 0. Then use Problem 38. To verify there exists a solution, you
could just differentiate the above formula using the fundamental theorem of calculus
and verify it works. Another way is to assume the solution in the form
x (t) = Φ (t) c (t)
and find c (t) to make it all work out. This is called the method of variation of
parameters.
For the method of variation of parameters,
x0 (t) =
AΦ (t) c (t) + Φ (t) c0 (t)
0
x (t) =
and so
c (t) =
AΦ (t) c (t) + f (t)
Z
t
−1
Φ (s)
f (s) ds + k
t0
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
84
Exercises
Then the solution desired is
x (t) = Φ (t) k + Φ (t)
Z
t
−1
Φ (s)
f (s) ds
t0
−1
Plug in t = t0 . This yields x0 = Φ (t0 ) k and so k = Φ (t0 ) x0 and so the solution is
Z t
−1
−1
x (t) = Φ (t) Φ (t0 ) x0 + Φ (t)
Φ (s) f (s) ds
t0
Note that by approximating with Riemann sums and passing to a limit, it follows that
the constant Φ (t) can be taken into the integral to write
Z t
−1
x (t) = Φ (t) Φ (t0 ) x0 +
Φ (t) Φ (s)−1 f (s) ds
t0
42. Show there exists a special Φ such that Φ0 (t) = AΦ (t) , Φ (0) = I, and suppose
−1
Φ (t) exists for all t. Show using uniqueness that
−1
Φ (−t) = Φ (t)
and that for all t, s ∈ R
Φ (t + s) = Φ (t) Φ (s)
Explain why with this special Φ, the solution to 6.35 can be written as
Z t
x (t) = Φ (t − t0 ) x0 +
Φ (t − s) f (s) ds.
t0
Hint: Let Φ (t) be such that the j th column is xj (t) where
x0j = Axj , xj (0) = ej .
Use uniqueness as required.
Just follow the hint. Such a Φ (t) clearly exists. By the Wronskian alternative in one
−1
of the above problems Φ (t) exists for all t.
Consider AΦ (t) − Φ (t) A = Ψ (t)
Ψ0 (t) =
=
=
AΦ0 (t) − Φ0 (t) A
AAΦ (t) − AΦ (t) A
A (Ψ (t))
and also Ψ (0) = 0. By uniqueness given above, Ψ (t) = 0 for all t. Therefore,
0
(Φ (t) Φ (−t))
= AΦ (t) Φ (−t) − Φ (t) AΦ (−t)
= Φ (t) AΦ (−t) − Φ (t) AΦ (−t) = 0
and so
Φ (t) Φ (−t) = C
a constant. However, when t = 0, C = I and so Φ (t) Φ (−t) = I. Also fixing s,
Ψ (t) ≡ Φ (t + s) − Φ (t) Φ (s)
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
85
Exercises
Ψ0 (t)
= Φ0 (t + s) − Φ0 (t) Φ (s)
= AΦ (t + s) − AΦ (t) Φ (s)
= AΨ (t)
and Ψ (0) = 0. Therefore, Ψ (t) = 0 for all t by uniqueness. It follows that for all t, s,
Φ (t + s) = Φ (t) Φ (s)
Therefore, in the variation of constants formula, it reduces to
Z t
x (t) = Φ (t − t0 ) x0 +
Φ (t − s) f (s) ds
t0
One thing might not be clear. If Ψ0 (t) = 0, why is Ψ (t) a constant? This works for
scalar valued functions by a use of the mean value theorem. However, this is a matrix
valued function. Nevertheless this is true. You just consider the ij th entry which is
eTi Ψ (t) ej
Its derivative equals 0 and so it is constant.
43. You can see more on this problem and the next one in the latest version of Horn
and Johnson, [16]. Two n × n matrices A, B are said to be congruent if there is an
invertible P such that
B = P AP ∗
Let A be a Hermitian matrix. Thus it has all real eigenvalues. Let n+ be the number
of positive eigenvalues, n− , the number of negative eigenvalues and n0 the number of
zero eigenvalues. For k a positive integer, let Ik denote the k × k identity matrix and
Ok the k × k zero matrix. Then the inertia matrix of A is the following block diagonal
n × n matrix.
In+
In−
On0
Show that A is congruent to its inertia matrix. Next show that congruence is an
equivalence relation. Finally, show that if two Hermitian matrices have the same
inertia matrix, then they must be congruent. Hint: First recall that there is a
unitary matrix, U such that
Dn+
Dn−
U ∗ AU =
On0
where the Dn+ is a diagonal
having the positive eigenvalues of A, Dn− being
matrix
defined similarly. Now let Dn− denote the diagonal matrix which replaces each entry
of Dn− with its absolute value. Consider the two diagonal matrices
−1/2
Dn+
Dn− −1/2
D = D∗ =
In0
Now consider D∗ U ∗ AU D.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
86
Exercises
The hint gives it away. When you block multiply, the thing which results is the inertia
matrix. Congruance is obviously an equivalence relation. Consider the transitive law
∗
for example. Say A = P BP ∗ , B = QCQ∗ . Then A = P QCQ∗ P ∗ = P QC (P Q) . Now
if you have two Hermitian matrices which have the same inertia matrix, then since
congruance is an equivalence relation, it follows that the two matrices are congruant.
44. Show that if A, B are two congruent Hermitian matrices, then they have the same
inertia matrix. Hint: Let A = SBS ∗ where S is invertible. Show that A, B have the
same rank and this implies that they are each unitarily similar to a diagonal matrix
which has the same number of zero entries on the main diagonal. Therefore, letting
VA be the span of the eigenvectors associated with positive eigenvalues of A and VB
being defined similarly, it suffices to show that these have the same dimensions. Show
that (Ax, x) > 0 for all x ∈ VA . Next consider S ∗ VA . For x ∈ VA , explain why
−1
(BS ∗ x,S ∗ x) =
S −1 A (S ∗ ) S ∗ x,S ∗ x
∗
= S −1 Ax,S ∗ x = Ax, S −1 S ∗ x = (Ax, x) > 0
Next explain why this shows that S ∗ VA is a subspace of VB and so the dimension of VB
is at least as large as the dimension of VA . Hence there are at least as many positive
eigenvalues for B as there are for A. Switching A, B you can turn the inequality
around. Thus the two have the same inertia matrix.
The above manipulation is straight forward. Now an orthonormal basis exists for Fn
which consists of eigenvectors of B. Say you have {v1 , · · · , vq , w1 , · · · , wr , z1 , · · · , zs }
where the vi correspond to the positive eigenvalues, the wi to the negative eigenvalues
and the zi to the eigenvalue 0. We know that B is positive on span (v1 , · · · , vq ) . Could
it be positive on any larger subspace? Could it be positive on span (v1 , · · · , vq , u)
where
u ∈ span (w1 , · · · , wr , z1 , · · · , zs )?
No, it couldn’t because you could then consider (Bu, u) and it would end up being no
larger than 0. It follows that S ∗ VA ⊆ VV = span (v1 , · · · , vq ) and so the dimension
of S ∗ VA , which is the same as the dimension of VA which is the same as the number
of positive eigenvalues, counted according to multiplicity, is no larger than q. Now
reverse the argument letting A ←→ B. Hence the two have the same number of
positive eigenvalues. Since they have the same number of zero eigenvalues, this implies
they have the same inertia matrix.
45. Let A be an m × n matrix. Then if you unraveled it, you could consider it as a vector
in Cnm . The Frobenius inner product on the vector space of m × n matrices is defined
as
(A, B) ≡ trace (AB ∗ )
Show that this really does satisfy the axioms of an inner product space and that it
also amounts to nothing more than considering m × n matrices as vectors in Cnm .
First note that
(A, B) =
X
Aij Bij =
i,j
(A, A) =
X
Aij Bij = (B, A)
i,j
X
i,j
Aij Aij =
X
i,j
|Aij |
2
Thus (A, A) = 0 if and only if A = 0. Clearly for a, b scalars,
(aA + bB, C) = a (A, C) + b (B, C)
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
87
Exercises
and so this does satisfy the axioms of the inner product. Also, this inner product
coincides with the standard inner product on Cnm .
46. ↑ Consider the n × n unitary matrices. Show that whenever U is such a matrix, it
follows that
√
|U |Cnn = n
Next explain why if {Uk } is any sequence of unitary matrices, there exists a subsequence {Ukm }∞
m=1 such that limm→∞ Ukm = U where U is unitary. Here the limit
takes place in the sense that the entries of Ukm converge to the corresponding entries
of U .
From the above and the fact that U is unitary,
2
|U |Cn2 = trace (U U ∗ ) = trace (I) = n.
2
Let {Uk } be a sequence of unitary matrices. Since these, considered as vectors in Cn ,
are contained in a closed and bounded set, it follows that there is a subsequence which
converges to U . Why is U a unitary matrix? You have
I = Ukm Uk∗m → U U ∗
and so U is unitary.
47. ↑Let A, B be two n × n matrices. Denote by σ (A) the set of eigenvalues of A. Define
dist (σ (A) , σ (B)) = max min {|λ − µ| : µ ∈ σ (B)}
λ∈σ(A)
Explain why dist (σ (A) , σ (B)) is small if and only if every eigenvalue of A is close
to some eigenvalue of B. Now prove the following theorem using the above problem
and Schur’s theorem. This theorem says roughly that if A is close to B then the
eigenvalues of A are close to those of B in the sense that every eigenvalue of A is close
to an eigenvalue of B.
Theorem F.24.2 Suppose limk→∞ Ak = A. Then
lim dist (σ (Ak ) , σ (A)) = 0
k→∞
By Schur’s theorem, there exist unitary matrices Uk such that
Uk∗ Ak Uk = Tk ,
where Tk is upper triangular. I claim that there exists a subsequence {Tkm }∞
m=1 such
that
Tkm → T
Where T is unitarily similar to A.
Using the above problem, there exists a subsequence km such that
lim Ukm = U
m→∞
a unitary matrix. It follows then that
lim Tkm = lim Uk∗m Akm Ukm = U ∗ AU ≡ T
n→∞
m→∞
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
88
Exercises
Since Uk∗m Akm Ukm is upper triangular, it follows that so is T . This shows the claim.
Next suppose that the conclusion does not hold. Then there exists ε > 0 and a
subsequence, still called k such that
dist (σ (Ak ) , σ (A)) ≥ ε
It follows that for every choice of Uk such that Uk∗ Ak Uk = Tk , a triangular matrix and
for every choice of unitary U such that U ∗ AU = T, a triangular matrix,
|Tk − T |Cn2 ≥ dist (σ (Ak ) , σ (A)) ≥ ε
Recall from the proof of Schur’s theorem, you change the order of the eigenvalues
on the diagonal by adjusting the unitary matrix U . No matter how you permute
the entries on the diagonals, the two are always far apart. Now pick Uk for each k,
such that Uk∗ Ak Uk = Tk a triangular matrix. By the first part of the proof, there
exists a subsequence {km } such that Tkm → T where T is unitarily similar to A. This
contradicts the above.
a b
48. Let A =
be a 2 × 2 matrix which is not a multiple of the identity. Show
c d
that A is similar to a 2 × 2 matrix which has one entry equal to 0. Hint: First note
that there exists a vector a such that Aa is not a multiple of a. Then consider
B=
a
Aa
−1
A
a Aa
Show B has a zero on the main diagonal.
p
p
Let the vector be
such that this vector and A
are linearly independent.
q
q
Then by brute force you do the following computation.
1
cp + dq −ap − bq
a b
p ap + bq
−q
p
c d
q cp + dq
D
Then you do the multiplication. This yields
1
0
(ad − bc) bq 2 − cp2 + apq − dpq cp2 − bq 2 − apq + dpq − (a + d) bq 2 − cp2 + apq − dpq
D
and since the trace equals 0 the term in the bottom right is also equal to 0.
49. ↑ Let A be a complex n × n matrix which has trace equal to 0. Show that A is similar
to a matrix which has all zeros on the main diagonal. Hint: Use Problem 30 on
Page 122 to argue that you can say that a given matrix is similar to one which has
the diagonal entries permuted in any order desired. Then use the above problem and
block multiplication to show that if the A has k nonzero entries, then it is similar to
a matrix which has k − 1 nonzero entries. Finally, when A is similar to one which has
at most one nonzero entry, this one must also be zero because of the condition on the
trace.
Suppose A has k nonzero diagonal entries. First of all, we can assume k ≥ 2 because
the trace equals 0. Then from Problem 30 on Page 122 it is similar to a block matrix
of the form
P Q
R S
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
89
Exercises
where the diagonal entries are just re ordered and P is a 2 × 2 matrix which which is
not a multiple of the identity and has both diagonal entries nonzero. Otherwise, the
trace could not equal zero if all the diagonal entries were equal. Then from the above
problem, there exists L such that L−1 P L has at least one zero on the diagonal.
−1
−1
L
0
P Q
L 0
L P L L−1 Q
=
0
I
R S
0 I
RL
S
Thus this new matrix is similar to the one above and it has at least one fewer nonzero
entries on the main diagonal. Continue this process obtaining a sequence of similar
matrices, each having fewer nonzero diagonal entries than the preceding one till there
is at most one nonzero entry. It follows that this one must be zero also because the
trace equals 0.
50. ↑An n × n matrix X is a comutator if there are n × n matrices A, B such that X =
AB − BA. Show that the trace of any comutator is 0. Next show that if a complex
matrix X has trace equal to 0, then it is in fact a comutator. Hint: Use the above
problem to show that it suffices to consider X having all zero entries on the main
diagonal. Then define
1
0
(
Xij
2
i−j if i 6= j
A=
, Bij =
.
..
0 if i = j
0
n
The first part is easy because the trace of a product is the same in either order.
Suppose then that X has zero trace. Then by the above problem, it is similar to a
matrix Y which has a zero diagonal. Suppose you can show that Y = AB − BA. Then
if Y = S −1 XS, you could write
S −1 XS = AB − BA,
X = S (AB − BA) S −1 = SAS −1 SBS −1 − SBS −1 SAS −1
Thus X will be a comutator. It suffices then to show that whenever a matrix has all
zero diagonal entries, it must be a comutator. Now use the definition of A, B in the
hint.
X
X
(AB − BA)ij ≡
Ais Bsj −
Bir Arj
s
r
Since A is a diagonal matrix, this reduces for i 6= j to
iBij − Bij j =
iXij
Xij
−j
= Xij
i−j
i−j
while if i = j, you have
iBii − Bii i = 0
Thus it yields Xij .
F.25
Exercises
8.4
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
90
Exercises
1
1
1
0
1. Let H denote span 2 , 4 , 3 , 1 . Find the dimension of H
0
0
1
1
and determine a basis.
1 0 0 −1
1 1 1 0
2 4 3 1 , row echelon form: 0 1 0 0 . The first three vectors form
0 0 1 1
0 0 1 1
a basis and the dimension is 3.
2. Let M = u = (u1 , u2 , u3 , u4 ) ∈ R4 : u3 = u1 = 0 . Is M a subspace? Explain.
Sure. It is closed with respect to linear combinations.
3. Let M = u = (u1 , u2 , u3 , u4 ) ∈ R4 : u3 ≥ u1 . Is M a subspace? Explain.
No. Not a subspace. Consider (0, 0, 1, 0) and multiply by −1.
4. Let w ∈ R4 and let M = u = (u1 , u2 , u3 , u4 ) ∈ R4 : w · u = 0 . Is M a subspace?
Explain.
Of course. It is the kernel of a linear transformation.
5. Let M = u = (u1 , u2 , u3 , u4 ) ∈ R4 : ui ≥ 0 for each i = 1, 2, 3, 4 . Is M a subspace?
Explain.
NO. Multiply something by −1.
6. Let w, w1 be given vectors in R4 and define
M = u = (u1 , u2 , u3 , u4 ) ∈ R4 : w · u = 0 and w1 · u = 0 .
Is M a subspace? Explain.
Yes. It is the intersection of two subspaces.
7. Let M = u = (u1 , u2 , u3 , u4 ) ∈ R4 : |u1 | ≤ 4 . Is M a subspace? Explain.
No. Take something nonzero in M where say u1 = 1. Now multiply by 100.
8. Let M = u = (u1 , u2 , u3 , u4 ) ∈ R4 : sin (u1 ) = 1 . Is M a subspace? Explain.
No. Let u1 = π/2 and multiply by 2.
9. Suppose {x1 , · · · , xk } is a set of vectors from Fn . Show that 0 is in span (x1 , · · · , xk ) .
P
0 = i 0xi
10. Consider the vectors of the form
2t + 3s
s − t : s, t ∈ R .
t+s
Is this set of vectors a subspace of R3 ? If so, explain why, give a basis for the subspace
and find its dimension.
2
3
Yes. A basis is −1 , 1
1
1
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
91
Exercises
11. Consider the vectors of the form
2t + 3s + u
s−t
t+s
u
: s, t, u ∈ R .
Is this set of vectors a subspace of R4 ? If so, explain why, give a basis for the subspace
and find its dimension.
It is a subspace. It is spanned by
3
1
1
0
It will be
3 2
1 1
1 1
0 0
pendent.
a basis if it is a linearly
1
0
, row echelon form:
0
1
2
1
1 0
, ,
1 0
0
1
independent set.
1 0 0
0 1 0
0 0 1 . Thus the vectors are linearly inde0 0 0
12. Consider the vectors of the form
2t + u + 1
t + 3u
t+s+v
u
: s, t, u, v ∈ R .
Is this set of vectors a subspace of R4 ? If so, explain why, give a basis for the subspace
and find its dimension.
Because of that 1 this is not a subspace.
13. Let V denote the set of functions defined on [0, 1]. Vector addition is defined as
(f + g) (x) ≡ f (x) + g (x) and scalar multiplication is defined as (αf ) (x) ≡ α (f (x)).
Verify V is a vector space. What is its dimension, finite or infinite?
Pick n points {x1 , · · · , xn } . Then let ei (x) = 0 unless x = xi when it equals 1. Then
n
{ei }i=1 is linearly independent, this for any n.
14. Let V denote the set of polynomial functions defined on [0, 1]. Vector addition is
defined as (f + g) (x) ≡ f (x) + g (x) and scalar multiplication is defined as (αf ) (x) ≡
α (f (x)). Verify V is a vector space. What is its dimension, finite or infinite?
This is clearly a subspace of the set of all functions
defined on
[0, 1] . Its dimension is
still infinite. For example, you can see that 1, x, x2 , · · · , xn is linearly independent
for each n. You could use Theorem 8.2.9 to show this, for example.
15. Let V be the set of polynomials defined on R having degree no more than 4. Give a
basis for this vector space.
1, x, x2 , x3 , x4
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
92
Exercises
√
16. Let the vectors be of the form a + b 2 where a, b are rational numbers and let the
field of scalars be F = Q, the rational numbers. Show directly this is a vector space.
What is its dimension? What is a basis for this vector space?
√ A basis is 1, 2 . In fact it is actually a field.
√
√ −1
2
a+b 2
= aa−b
2 −2b2 . Since a, b are rational, it follows that the denominator is not
0. Thus every element has an inverse so it is actually a field. It is certainly a vector
space over the rational numbers.
17. Let V be a vector space with field of scalars F and suppose {v1 , · · · , vn } is a basis for
V . Now let W also be a vector space with field of scalars F. Let L : {v1 , · · · , vn } →
W be a function such that Lvj = wj . Explain how L can be extended to a linear
transformation mapping V to W in a unique way.
Pn
Pn
L ( i=1 ci vi ) ≡ i=1 ci wi
18. If you have 5 vectors in F5 and the vectors are linearly independent, can it always be
concluded they span F5 ? Explain.
Yes. If not, you could add in a vector not in the span and obtain a linearly independent
set of 6 vectors which is not possible.
19. If you have 6 vectors in F5 , is it possible they are linearly independent? Explain.
No. There is a spanning set having 5 vectors and this would need to be as long as the
linearly independent set.
20. Suppose V, W are subspaces of Fn . Show V ∩ W defined to be all vectors which are in
both V and W is a subspace also.
If x, y ∈ V ∩ W then αa + βb ∈ V and is also in W because these are subspaces. Hence
V ∩ W is a subspace.
21. Suppose V and W both have dimension equal to 7 and they are subspaces of a vector
space of dimension 10. What are the possibilities for the dimension of V ∩ W ? Hint:
Remember that a linear independent set can be extended to form a basis.
See the next problem.
22. Suppose V has dimension p and W has dimension q and they are each contained in
a subspace, U which has dimension equal to n where n > max (p, q) . What are the
possibilities for the dimension of V ∩ W ? Hint: Remember that a linear independent
set can be extended to form a basis.
Let {x1 , · · · , xk } be a basis for V ∩ W. Then there is a basis for V and W which are
respectively
{x1 , · · · , xk , yk+1 , · · · , yp } , {x1 , · · · , xk , zk+1 , · · · , zq }
It follows that you must have k + p − k + q − k ≤ n and so you must have
p+q−n≤k
23. If b 6= 0, can the solution set of Ax = b be a plane through the origin? Explain.
No. It can’t. It does not contain 0.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
Exercises
93
24. Suppose a system of equations has fewer equations than variables and you have found
a solution to this system of equations. Is it possible that your solution is the only one?
Explain.
No. There must then be infinitely many solutions. If the system is Ax = b, then
there are infinitely many solutions to Ax = 0 and so the solutions to Ax = b are a
particular solution to Ax = b added to the solutions to Ax = 0 of which there are
infinitely many.
25. Suppose a system of linear equations has a 2×4 augmented matrix and the last column
is a pivot column. Could the system of linear equations be consistent? Explain.
No. This would lead to 0 = 1.
26. Suppose the coefficient matrix of a system of n equations with n variables has the
property that every column is a pivot column. Does it follow that the system of
equations must have a solution? If so, must the solution be unique? Explain.
Yes. It has a unique solution.
27. Suppose there is a unique solution to a system of linear equations. What must be true
of the pivot columns in the augmented matrix.
The last one must not be a pivot column and the ones to the left must each be pivot
columns.
28. State whether each of the following sets of data are possible for the matrix equation
Ax = b. If possible, describe the solution set. That is, tell whether there exists a
unique solution no solution or infinitely many solutions.
(a) A is a 5 × 6 matrix, rank (A) = 4 and rank (A|b) = 4. Hint: This says b is in
the span of four of the columns. Thus the columns are not independent.
Infinite solution set.
(b) A is a 3 × 4 matrix, rank (A) = 3 and rank (A|b) = 2.
This surely can’t happen. If you add in another column, the rank does not get
smaller.
(c) A is a 4 × 2 matrix, rank (A) = 4 and rank (A|b) = 4. Hint: This says b is in
the span of the columns and the columns must be independent.
You can’t have the rank equal 4 if you only have two columns.
(d) A is a 5 × 5 matrix, rank (A) = 4 and rank (A|b) = 5. Hint: This says b is not
in the span of the columns.
In this case, there is no solution to the system of equations represented by the
augmented matrix.
(e) A is a 4 × 2 matrix, rank (A) = 2 and rank (A|b) = 2.
In this case, there is a unique solution since the columns of A are independent.
29. Suppose A is an m × n matrix in which m ≤ n. Suppose also that the rank of A equals
m. Show that A maps Fn onto Fm . Hint: The vectors e1 , · · · , em occur as columns
in the row reduced echelon form for A.
This says that the columns of A have a subset of m vectors which are linearly independent. Therefore, this set of vectors is a basis for Fm . It follows that the span of
the columns is all of Fm . Thus A is onto.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
94
Exercises
30. Suppose A is an m × n matrix in which m ≥ n. Suppose also that the rank of A equals
n. Show that A is one to one. Hint: If not, there exists a vector, x such that Ax = 0,
and this implies at least one column of A is a linear combination of the others. Show
this would require the column rank to be less than n.
The columns are independent. Therefore, A is one to one.
31. Explain why an n × n matrix A is both one to one and onto if and only if its rank is
n.
The rank is n is the same as saying the columns are independent which is the same as
saying A is one to one which is the same as saying the columns are a basis. Thus the
span of the columns of A is all of Fn and so A is onto. If A is onto, then the columns
must be linearly independent since otherwise the span of these columns would have
dimension less than n and so the dimension of Fn would be less than n.
32. If you have not done this problem already, here it is again. It is a very important
result. Suppose A is an m × n matrix and B is an n × p matrix. Show that
dim (ker (AB)) ≤ dim (ker (A)) + dim (ker (B)) .
Let {w1 , · · · , wk } be a basis for B (Fp ) ∩ ker (A) and suppose {u1 , · · · , ur } is a basis
for ker (B). Let Bzi = wi . Now suppose x ∈ ker (AB). Then Bx ∈ ker (A) ∩ B (Fp )
and so
k
k
X
X
Bx =
a i wi =
ai Bzi
i=1
so
x−
i=1
k
X
i=1
Hence
x−
k
X
ai zi ∈ ker (B)
ai zi =
i=1
This shows that
r
X
b j uj
j=1
dim (ker (AB)) ≤ k + r ≤ dim (ker (A)) + dim (ker (B))
Note that a little more can be said. {u1 , · · · , ur , z1 , · · · , zk } is independent. Here is
why. Suppose
k
X
X
a i ui +
bj zj = 0
i
j=1
Then do B to both sides and conclude that
X
bj Bzj = 0
j
which forces each bj = 0. Then the independence of the ui implies each ai = 0. It
follows that in fact,
dim (ker (AB)) = k + r ≤ dim (ker (A)) + dim (ker (B))
and the remaining inequality is an equality if and only if B (Fp ) ⊇ ker (A).
This is because B (Fp ) ∩ ker (A) ⊆ ker (A) and these are equal exactly when they have
the same dimension.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
95
Exercises
33. Recall that every positive integer can be factored into a product of primes in a unique
way. Show there must be infinitely many primes. Hint: Show that if you have any
finite set of primes and you multiply them and then add 1, the result cannot be
divisible by any of the primes in your finite set. This idea in the hint is due to Euclid
who lived about 300 B.C.
Consider p1 p2 · · · pn + 1. Then if
p1 p2 · · · pn + 1 = lpk
you would have
1
+m=l
pk
where m is an integer, the product of the other primes different than pk . This is a
contradiction. Hence there must be another prime to divide p1 p2 · · · pn + 1 and so
there are infinitely many primes.
34. There are lots of fields. This will give an example of a finite field. Let Z denote the set
of integers. Thus Z = {· · · , −3, −2, −1, 0, 1, 2, 3, · · ·}. Also let p be a prime number.
We will say that two integers, a, b are equivalent and write a ∼ b if a − b is divisible
by p. Thus they are equivalent if a − b = px for some integer x. First show that
a ∼ a. Next show that if a ∼ b then b ∼ a. Finally show that if a ∼ b and b ∼ c
then a ∼ c. For a an integer, denote by [a] the set of all integers which is equivalent
to a, the equivalence class of a. Show first that is suffices to consider only [a] for
a = 0, 1, 2, · · · , p − 1 and that for 0 ≤ a < b ≤ p − 1, [a] 6= [b]. That is, [a] = [r] where
r ∈ {0, 1, 2, · · · , p − 1}. Thus there are exactly p of these equivalence classes. Hint:
Recall the Euclidean algorithm. For a > 0, a = mp + r where r < p. Next define the
following operations.
[a] + [b] ≡ [a + b]
[a] [b] ≡ [ab]
Show these operations are well defined. That is, if [a] = [a0 ] and [b] = [b0 ] , then
[a] + [b] = [a0 ] + [b0 ] with a similar conclusion holding for multiplication. Thus for
addition you need to verify [a + b] = [a0 + b0 ] and for multiplication you need to verify
[ab] = [a0 b0 ]. For example, if p = 5 you have [3] = [8] and [2] = [7] . Is [2 × 3] = [8 × 7]?
Is [2 + 3] = [8 + 7]? Clearly so in this example because when you subtract, the result
is divisible by 5. So why is this so in general? Now verify that {[0] , [1] , · · · , [p − 1]}
with these operations is a Field. This is called the integers modulo a prime and is
written Zp . Since there are infinitely many primes p, it follows there are infinitely
many of these finite fields. Hint: Most of the axioms are easy once you have shown
the operations are well defined. The only two which are tricky are the ones which
give the existence of the additive inverse and the multiplicative inverse. Of these, the
first is not hard. − [x] = [−x]. Since p is prime, there exist integers x, y such that
1 = px+ky and so 1−ky = px which says 1 ∼ ky and so [1] = [ky] . Now you finish the
argument. What is the multiplicative identity in this collection of equivalence classes?
Of course you could now consider field extensions based on these fields.
The only substantive issue is why Zp is a field. Let [x] ∈ Zp where [x] 6= [0]. Thus x
is not a multiple of p. Then x and p are relatively prime. Hence from Theorem 1.9.3,
there exists a, b such that
1 = ap + bx
Then
[1 − bx] = [ap] = 0
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
96
Exercises
and it follows that
[b] [x] = [1]
so [b] = [x]
−1
.
35. Suppose the field of scalars is Z2 described above. Show that
0 1
0 0
0 0
0 1
1
−
=
0 0
1 0
1 0
0 0
0
0
1
Thus the identity is a comutator. Compare this with Problem 50 on Page 198.
1 0
1 0
This is easy. The left side equals
=
0 −1
0 1
36. Suppose V is a vector space with field of scalars F. Let T ∈ L (V, W ) , the space of
linear transformations mapping V onto W where W is another vector space. Define
an equivalence relation on V as follows. v ∼ w means v − w ∈ ker (T ) . Recall that
ker (T ) ≡ {v : T v = 0}. Show this is an equivalence relation. Now for [v] an equivalence class define T 0 [v] ≡ T v. Show this is well defined. Also show that with the
operations
[v] + [w]
α [v]
≡ [v + w]
≡ [αv]
this set of equivalence classes, denoted by V / ker (T ) is a vector space. Show next that
T 0 : V / ker (T ) → W is one to one, linear, and onto. This new vector space, V / ker (T )
is called a quotient space. Show its dimension equals the difference between the
dimension of V and the dimension of ker (T ).
It is obviously an equivalence relation. Is the operation of addition well defined? If
[v] = [v0 ] , similar for w, w0 , is it the case that [v + w] = [v0 + w0 ]? Is
v + w− (v0 + w0 ) ∈ ker (T )?
Of course so. It is just the sum of two things in ker (T ) . Similarly the operation of
scalar multiplication is well defined. Is T 0 well defined? If v ∼ w, is it true that
T v =T w? Of course so.
T (v − w) = 0
Is T 0 linear? This is obvious. Is T 0 one to one? If T 0 [v] = 0, then it says that T v = 0
and so v ∈ ker (T ) . Hence [v] = [0] . Thus T 0 is certainly one to one. If T is onto, it is
obvious that T 0 is onto. Just pick w ∈ W . Then T v = w for some v. Then T 0 [v] = w.
37. Let V be an n dimensional vector space and let W be a subspace. Generalize the
above problem to define and give properties of V /W . What is its dimension? What
is a basis?
Here you define an equivalence relation by v ∼ u if v − u ∈ W. This is clearly
an equivalence relation. Define [v] + [u] ≡ [u + v] and α [u] ≡ [αu]. Then by a
repeat of the above, everything is well defined. It is clear that V /W , the set of
equivalence classes is a vector space with these operations as just defined. What is
its dimension? Let {w1 , · · · , wr } be a basis for W . Now extend to get a basis for
V, {w1 , · · · , wr , v1 , · · · , vn−r }. Then if
n−r
X
ci [vi ] = [0] ,
i=1
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
97
Exercises
it follows that
Pn−r
i=1
ci vi ∈ W and so there exist scalars di such that
r
X
di wi +
i=1
n−r
X
ci vi = 0
i=1
Now, since {w1 , · · · , wr , v1 , · · · , vn−r } is a basis, it follows that all the di and ci are
equal to 0. Therefore, in particular the vectors [vi ] form a linearly independent set in
V /W. Do these vectors span V /W ? Suppose [v] ∈ V /W . Then
v=
r
X
di wi +
i=1
n−r
X
ci vi
i=1
for suitable scalars ci and di . Now it follows that
[v] =
n−r
X
ci [vi ]
i=1
and so the dimension of V /W equals n − r where r is the dimension of W .
38. If F and G are two fields and F ⊆ G, can you consider G as a vector space with field
of scalars F? Explain.
Yes. This is obvious.
39. Let A denote the algebraic numbers, those numbers which are roots of polynomials
having rational coefficients which are in R. Show A can be considered a vector space
with field of scalars Q. What is the dimension of this vector space, finite or infinite?
As in the above problem, since A is a field, it can be considered as a vector space
over Q. So what is the dimension? It is clearly infinite. To see this, consider xn − 7.
By the rational √
root theorem, this polynomial
√ is irreducible. Therefore, the minimal
polynomial for n 7 is xn − 7 and so for α = n 7, 1, α, α2 , · · · , αn−1 are linearly independent since otherwise, xn − 7 would not be the minimal polynomial of α and would
end up not being irreducible. Since n is arbitrary, this shows that the dimension of A
is infinite.
40. As mentioned, for distinct algebraic numbers αi , the complex numbers {eαi }ni=1 are
linearly independent over the field of scalars A where A denotes the algebraic numbers,
those which are roots of a polynomial having integer (rational) coefficients. What is
the dimension of the vector space C with field of scalars A, finite or infinite? If the
field of scalars were C instead of A, would this change? What if the field of scalars
were R?
How many distinct algebraic numbers can you get? Clearly as many as desired. Just
consider the nth roots of some complex number. If you have n distinct algebraic
n
numbers αi , then from the Lindemann Weierstrass theorem mentioned above, {eαi }i=1
is linearly independent. Of course if the field of scalars were C this would change
completely. Then the dimension of C over C is clearly 1.
41. Suppose F is a countable field and let A be the algebraic numbers, those numbers
which are roots of a polynomial having coefficients in F which are in G, some other
field containing F. Show A is also countable.
You consider each polynomial of degree n which has coefficients in the countable field
F. There are countably many of these. Each has finitely many roots in A. Therefore,
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
98
Exercises
there are countably many things in A which correspond to polynomials of degree n
The totality of A is obtained then from taking a countable union of countable sets,
and this is countable.
42. This problem is on partial fractions. Suppose you have
R (x) =
p (x)
, degree of p (x) < degree of denominator.
q1 (x) · · · qm (x)
where the polynomials qi (x) are relatively prime and all the polynomials p (x) and
qi (x) have coefficients in a field of scalars F. Thus there exist polynomials ai (x)
having coefficients in F such that
1=
m
X
ai (x) qi (x)
i=1
Explain why
P
m
X
p (x) m
a (x) p (x)
i=1 ai (x) qi (x)
Qi
R (x) =
=
q1 (x) · · · qm (x)
j6=i qj (x)
i=1
This is because you just did
Pm
p (x) i=1 ai (x) qi (x)
p (x)
R (x) =
=
q1 (x) · · · qm (x)
q1 (x) · · · qm (x)
Then simplifying this yields
R (x) =
m
X
a (x) p (x)
Qi
j6=i qj (x)
i=1
Now continue doing this on each term in the above sum till finally you obtain an
expression of the form
m
X
bi (x)
i=1
qi (x)
Using the Euclidean algorithm for polynomials, explain why the above is of the form
M (x) +
m
X
ri (x)
i=1
qi (x)
where the degree of each ri (x) is less than the degree of qi (x) and M (x) is a polynomial.
You just note that
bi (x) = li (x) qi (x) + ri (x)
where the degree of ri (x) is less than the degree of qi (x) . Then replacing each bi (x)
by this expression, you obtain the above expression.
Now argue that M (x) = 0.
You have
M (x) +
m
X
ri (x)
i=1
qi (x)
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
=
p (x)
q1 (x) · · · qm (x)
7KH6D\ORU)RXQGDWLRQ
99
Exercises
Now multiply both sides by q1 (x) · · · qm (x) . If M (x) is not zero, then you would have
the degree of the left is larger than the degree on the right in the equation which
results. Hence M (x) = 0.
From this explain why the usual partial fractions expansion of calculus must be true.
You can use the fact that every polynomial having real coefficients factors into a product of irreducible quadratic polynomials and linear polynomials having real coefficients.
This follows from the fundamental theorem of algebra in the appendix.
n
What if some qi (x) = ki (x) ? In this case, you can go further. You have the partial
fractions expansion described above and you have a term of the form
a (x)
n
n , degree of a (x) < degree k (x)
k (x)
I claim this can be written as
n
X
ai (x)
i=1
k (x)
i
where the degree of ai (x) < degree of k (x). This can be proved by induction. If
n = 1, there is nothing to show. Suppose then it is true for n − 1.
!
1
a (x)
a (x)
=
k (x)n
k (x) k (x)n−1
n−1
If the degree of a (x) is less than the degree of k (x)
, then there is nothing to show.
Induction takes care of it. If this is not the case, you can write
!
!
n−1
a (x)
1
a (x)
1
b (x) k (x)
+ r (x)
=
n =
n−1
k (x) k (x)n−1
k (x)
k (x)
k (x)
where the degree of r (x) is less than the degree of k (x)
n−1
. Then this reduces to
b (x)
a (x)
r (x)
1
+
n =
k (x) k (x)n−1 k (x)
k (x)
The second term gives what is desired by induction. I claim that the degree of b (x)
n
must be less than the degree of k (x) . If not, you could multiply both sides by k (x)
and obtain
a (x) = b (x) k (x)n−1 + r (x)
n
and the degree of the left is less than the degree of k (x) while the degree of the
n
right is greater than the degree of k (x) . Therefore, the desired decomposition has
been obtained. Now the usual procedure for partial fractions follows. If you have
a rational function p (x) /q (x) where the degree of q (x) is larger than the degree
of p (x) , there exist factors of q (x) into products of linear and irreducible over R
quadratic factors by using the fundamental theorem of algebra. Then you just apply
the above decomposition result.
43. Suppose {f1 , · · · , fn } is an independent set of smooth functions defined on some interval (a, b). Now let A be an invertible n × n matrix. Define new functions {g1 , · · · , gn }
as follows.
g1
f1
..
.
. = A ..
gn
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
fn
7KH6D\ORU)RXQGDWLRQ
100
Exercises
Is it the case that {g1 , · · · , gn } is also independent? Explain why.
P
Suppose ni=1 ai gi = 0. Then
0=
X
i
ai
X
Aij fj =
j
It follows that
X
X
j
fj
X
Aij ai
i
Aij ai = 0
i
for each j. Therefore, since AT is invertible, it follows that each ai = 0. Hence the
functions gi are linearly independent.
F.26
Exercises
9.5
1. If A, B, and C are each n × n matrices and ABC is invertible, why are each of A, B,
and C invertible?
This is because ABC is one to one.
2. Give an example of a 3 × 2 matrix with the property that the linear transformation
determined by this matrix is one to one but not onto.
1 0
0 1
0 0
3. Explain why Ax = 0 always has a solution whenever A is a linear transformation.
4. Review problem: Suppose det (A − λI) = 0. Show using Theorem 3.1.15 there exists
x 6= 0 such that (A − λI) x = 0.
Since the matrix (A − λI) is not invertible, it follows that there exists x which it sends
to 0.
5. How does the minimal polynomial of an algebraic number relate to the minimal polynomial of a linear transformation? Can an algebraic number be thought of as a linear
transformation? How?
They are pretty much the same thing. If you have an algebraic number α it can
be thought of as a linear transformation which maps A to A according to the rule
α (β) 6= αβ.
6. Recall the fact from algebra that if p (λ) and q (λ) are polynomials, then there exists
l (λ) , a polynomial such that
q (λ) = p (λ) l (λ) + r (λ)
where the degree of r (λ) is less than the degree of p (λ) or else r (λ) = 0. With this in
mind, why must the minimal polynomial always divide the characteristic polynomial?
That is, why does there always exist a polynomial l (λ) such that p (λ) l (λ) = q (λ)?
Can you give conditions which imply the minimal polynomial equals the characteristic
polynomial? Go ahead and use the Cayley Hamilton theorem.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
101
Exercises
If q (λ) is the characteristic polynomial and p (λ) is the minimal polynomial, then from
the above equation,
0 = r (A)
and this cannot happen unless r (λ) = 0 because if not, p (λ) was not the minimal
polynomial after all.
7. In the following examples, a linear transformation, T is given by specifying its action
on a basis β. Find its matrix with respect to this basis.
1
1
−1
−1
−1
(a) T
=2
+1
,T
=
2
2
1
1
1
1
−1
The basis is
,
. Formally,
2
1
1
−1
−1
1
−1
2
+1
,
=
,
A
2
1
1
2
1
1 −1
1 −1
=
A
5 1
2 1
−1 1 −1
1 −1
2 0
A=
=
2 1
5 1
1 1
0
0
−1
−1
0
(b) T
=2
+1
,T
=
1
1
1
1
1
0
−1
0
0
−1
2
+1
,
=
,
A
1
1
1
1
1
2 1
A=
1 0
1
1
1
1
1
1
(c) T
=2
+1
,T
=1
−
0
2
0
2
0
2
1
1
1
1
1
1
2
+1
,1
−
=
,
A
2
0
0
2
0
2
1 1
A=
2 −1
8. Let β = {u1 , · · · , un } be a basis for Fn and let T : Fn → Fn be defined as follows.
!
n
n
X
X
T
a k uk =
a k b k uk
k=1
k=1
First show that T is a linear transformation. Next show that the matrix of T with
respect to this basis is [T ]β =
b1
..
.
bn
Show that the above definition is equivalent to simply specifying T on the basis vectors
of β by
T (uk ) = bk uk .
Formally,
(b1 u1 , b2 u2 , · · · , bn un ) = (u1 , u2 , · · · , un ) [T ]β
Thus [T ]β is the diagonal matrix just described.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
102
Exercises
9. ↑In the situation of the above problem, let γ = {e1 , · · · , en } be the standard basis for
Fn where ek is the vector which has 1 in the k th entry and zeros elsewhere. Show that
[T ]γ =
−1
u1 · · · un [T ]β u1 · · · un
(6.36)
We have T ui = bi ui and so we know [T ]β . The problem is to find [T ]γ where γ is the
good basis. Consider the diagram
Fn
↓
Fn
↑
Fn
[T ]γ
→
T
→
[T ]β
→
Fn
↓
Fn
↑
Fn
−1
u1 · · · un
On the left side going down you get x → x →
x.
Going up on the
right, you need to have x → u1 · · · un x → u1 · · · un x. Then you need
[T ]γ =
u1
···
un
[T ]β
u1
···
un
−1
10. ↑Generalize the above problem to the situation where T is given by specifying its
action on the vectors of a basis β = {u1 , · · · , un } as follows.
T uk =
n
X
ajk uj .
j=1
Letting A = (aij ) , verify that for γ = {e1 , · · · , en } , 6.36 still holds and that [T ]β = A.
This isn’t really any different. As explained earlier, the matrix of T with respect to
the given basis is just A where it is as described. Thus A = [T ]β and by a repeat of
the above argument,
−1
[T ]γ = u1 · · · un [T ]β u1 · · · un
11. Let P3 denote the set of real polynomials of degree no more than 3, defined on an
interval [a, b]. Show that P3 is a subspace of the vector space
of all functions defined
on this interval. Show that a basis for P3 is 1, x, x2 , x3 . Now let D denote the
differentiation operator which sends a function to its derivative. Show D is a linear
transformation which sends P3 to P3 . Find the matrix of this linear transformation
with respect to the given basis.
0, 1, 2x, 3x2 = 1, x, x2 , x3 A where A is the desired matrix. Then clearly
0 1 0 0
0 0 2 0
A=
0 0 0 3
0 0 0 0
12. Generalize the above problem to Pn , the space of polynomials of degree no more than
n with basis {1, x, · · · , xn } .
The pattern seems pretty clear. A will be an (n + 1) × (n + 1) matrix which has zeros
everywhere except on the super diagonal where you have, starting at the top and going
towards the bottom, the numbers in the following order. 1, 2, 3, · · · , n.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
103
Exercises
13. In the situation of the above problem, let the linear transformation be T = D2 + 1,
defined as T f = f 00 + f. Find the matrix of this linear transformation with respect to
the given basis {1, x, · · · , xn }.
First consider the operator D2 .
0 0 2 6x 12x2 · · · =
1
x x2
To see the pattern, let n = 4. Then you
0
0
A=
0
0
0
Then the matrix will be this one added
1 0
0 1
0 0
0 0
0 0
x3
x4
would have
0
0
0
0
0
2
0
0
0
0
0
6
0
0
0
0
0
12
0
0
···
A
to the identity. Hence it is the matrix
2 0 0
0 6 0
1 0 12
0 1 0
0 0 1
14. In calculus, the following situation is encountered. There exists a vector valued function f :U → Rm where U is an open subset of Rn . Such a function is said to have
a derivative or to be differentiable at x ∈ U if there exists a linear transformation
T : Rn → Rm such that
lim
v→0
|f (x + v) − f (x) − T v|
= 0.
|v|
First show that this linear transformation, if it exists, must be unique. Next show
that for β = {e1 , · · · , en } , , the standard basis, the k th column of [T ]β is
∂f
(x) .
∂xk
Actually, the result of this problem is a well kept secret. People typically don’t see
this in calculus. It is seen for the first time in advanced calculus if then.
Suppose that T 0 also works. Then letting t > 0,
|f (x + tv) − f (x) − tT v|
|f (x + tv) − f (x) − tT 0 v|
= lim
=0
t→0
t→0
t
t
lim
In particular,
t (T v − T 0 v)
=0
t→0
t
which says that T v = T 0 v and since v is arbitrary, this shows T = T 0 .
lim
15. Recall that A is similar to B if there exists a matrix P such that A = P −1 BP. Show
that if A and B are similar, then they have the same determinant. Give an example
of two matrices which are not similar but have the same determinant.
These two are not similar but have the same determinant.
1 0
1 1
,
0 1
0 1
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
104
Exercises
You can see these are not similar by noticing that the second has an eigenspace of
dimension equal to 1 so it is not similar to any diagonal matrix which is what the first
one is.
16. Suppose A ∈ L (V, W ) where dim (V ) > dim (W ) . Show ker (A) 6= {0}. That is, show
there exist nonzero vectors v ∈ V such that Av = 0.
If not, then A would be one to one and so it would take a basis to a linearly independent
set which is impossible because W has smaller dimension than V .
17. A vector v is in the convex hull of a nonempty set S if there are finitely many vectors
of S, {v1 , · · · , vm } and nonnegative scalars {t1 , · · · , tm } such that
v=
m
X
tk vk ,
k=1
m
X
tk = 1.
k=1
Such a linear combination is called a convex combination.
Suppose now that S ⊆ V,
Pm
a vector space of dimension n. Show that if v = k=1 tk vk is a vector in the convex
hull for m > n + 1, then there exist other scalars {t0k } such that
v=
m−1
X
t0k vk .
k=1
Thus every vector in the convex hull of S can be obtained as a convex combination
of at most n + 1 points of S. This incredible result is in Rudin [23]. Hint: Consider
L : Rm → V × R defined by
!
m
m
X
X
L (a) ≡
ak vk ,
ak
k=1
k=1
Explain why ker (L) 6= {0} . Next, letting a ∈ ker (L) \ {0} and λ ∈ R, note that
λa ∈ ker (L) . Thus for all λ ∈ R,
v=
m
X
(tk + λak ) vk .
k=1
Now vary λ till some tk + λak = 0 for some ak 6= 0.
If any of the tk = 0, there is nothing to prove. Hence you can assume each of these
tk > 0. Suppose then that m > n + 1 and consider
!
m
m
X
X
L (a) ≡
ak vk ,
ak
k=1
k=1
m
a mapping from R to V × R a vector space of dimension n + 1. Then since m > n + 1,
this mapping L cannot be one to one. Hence
Pmthere exists a 6= 0 such that La = 0.
Thus also λa ∈ ker (L). By assumption, v = k=1 tk vk . Then for each λ ∈ R,
v=
m
X
(tk + λak ) vk
k=1
P
P
P
(tk + λak ) = 1 because
ak = 0. Also, for λ = 0,
because m
k=1 ak vk = 0. Now
each of the tk + λak > 0. Adjust λ so that each of these is still ≥ 0 but one vanishes.
Then you have gotten v as a convex combination of fewer than m vectors.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
105
Exercises
18. For those who know about compactness, use Problem 17 to show that if S ⊆ Rn and
S is compact, then so is its convex hull.
Suppose that the convex hull of S is not compact. Then nthere exists
o
∞
Pn+1
v1k , v2k , · · · , v(n+1)k k=1 , each in S and sk ∈ C ≡ s : i=1 si = 1, si ∈ [0, 1]
such that
(n+1
)∞
X
sik vik
i=1
k=1
has no convergent subsequence. However, there exists a further subsequence, still
denoted as k such that for each i, limk→∞ vik = vi . This is because S is given to be
compact. Also, taking yet another subsequence, you can assume that limk→∞ sik = si
because of the compactness of C. But this yields an obvious contradiction to the
assertion that the above has no convergent subsequence.
19. Suppose Ax = b has a solution. Explain why the solution is unique precisely when
Ax = 0 has only the trivial (zero) solution.
This is because the general solution is yp +y where Ayp = b and Ay = 0. Now A0 = 0
and so the solution is unique precisely when this is the only solution y to Ay = 0.
20. Let A be an n × n matrix of elements of F. There are two cases. In the first case,
F contains a splitting field of pA (λ) so that p (λ) factors into a product of linear
polynomials having coefficients in F. It is the second case which of interest here where
pA (λ) does not factor into linear factors having coefficients in F. Let G be a splitting
field of pA (λ) and let qA (λ) be the minimal polynomial of A with respect to the field
G. Explain why qA (λ) must divide pA (λ). Now why must qA (λ) factor completely
into linear factors?
In the field G
pA (λ) = qA (λ) l (λ) + r (λ)
where r (λ) either equals 0 or has degree less than the degree of qA (λ). Then it
follows that r (A) = 0 and so it must be the case that r (λ) = 0. Hence qA (λ)
must divide pA (λ). Now G was the splitting field of pA (λ) and this new polynomial
divides this one. Therefore, this new polynomial also must factor completely into
linear polynomials having coefficients in G.
pA (λ) = qA (λ) l (λ)
and all roots of pA (λ) are in G so the same is true of the roots of qA (λ) because these
are all roots of pA (λ).
21. In Lemma 9.2.2 verify that L is linear.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
106
Exercises
L a
= L
n
X
αi vi + b
i=1
n
X
n
X
(aαi + bβ i ) vi
i=1
=
n
X
= a
αi Lvi + b
i=1
n
X
β i Lvi
i=1
n
X
i=1
F.27
!
(aαi + bβ i ) Lvi
i=1
n
X
= aL
β i vi
i=1
!
αi vi
!
+ bL
n
X
β i vi
i=1
!
Exercises
10.6
1. In the discussion of Nilpotent transformations, it was asserted that if two n×n matrices
A, B are similar, then Ak is also similar to B k . Why is this so? If two matrices are
similar, why must they have the same rank?
Say A = S −1 BS. Then Ak = S −1 B k S and so these are also similar. You can write
S −1 BSS −1 BSS −1 BS · · · S −1 BS
and note that the inside SS −1 cancels. Therefore, the result of the above is as claimed.
Hence Ak is similar to B k .
Now the dimension of A (Fn ) is the dimension of S −1 BS (Fn ) which is the same as the
dimension of S −1 B (Fn ) which is the same as the dimension of B (Fn ) because S, S −1
are one to one and onto. Hence they each take a basis to a basis.
2. If A, B are both invertible, then they are both row equivalent to the identity matrix.
Are they necessarily similar? Explain.
1 1
1 0
Obviously not. Consider
,
. These are both in Jordan form.
0 1
0 1
3. Suppose you have two nilpotent matrices A, B and Ak and B k both have the same
rank for all k. Does it follow that A, B are similar? What if it is not known that A, B
are nilpotent? Does it follow then?
Yes, this is so. You have A = S −1 JS and B = T −1 J 0 T where J, J 0 are both Jordan
canonical form. Then you have that J = SAS −1 and J 0 = T BT −1 and so J k and J 0k
have the same rank for all k. Therefore, J = J 0 because these matrices are nilpotent.
It follows that
SAS −1 = T BT −1
and consequently,
"
−1
B = T −1 SAS −1 T = S −1 T
AS −1 T
so that A is similar to B. In case they are not nilpotent, the above problem gives a
counter example.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
107
Exercises
4. When we say a polynomial equals zero, we mean that all the coefficients equal 0. If
we assign a different meaning to it which says that a polynomial
p (λ) =
n
X
ak λk = 0,
k=0
when the value of the polynomial equals zero whenever a particular value of λ ∈ F
is placed in the formula for p (λ) , can the same conclusion be drawn? Is there any
difference in the two definitions for ordinary fields like Q? Hint: Consider Z2 , the
integers mod 2.
This sort of thing where it equals 0 if it sends everything to 0 works fine when we
think of polynomials as functions. However, consider λ2 + λ where the coefficients
come from Z2 . It sends everything to 0 but we don’t want to consider this polynomial
to be the zero polynomial because its coefficients are not all 0.
5. Let A ∈ L (V, V ) where V is a finite dimensional vector space with field of scalars F.
Let p (λ) be the minimal polynomial and suppose φ (λ) is any nonzero polynomial such
that φ (A) is not one to one and φ (λ) has smallest possible degree such that φ (A) is
nonzero and not one to one. Show φ (λ) must divide p (λ).
Say φ (A) x = 0 where x 6= 0. Consider all monic polynomials φ (λ) such that φ (A) x = 0.
One such is the minimal polynomial p (λ). This condition is less severe than saying
that φ (A) = 0. Therefore, there are more polynomials considered. If you take the one
of least degree, its degree must be no larger than the degree of p (λ) . By assumption
there is a polynomial φ (λ) for which φ (A) 6= 0 but such that φ (A) x = 0. You can do
the usual thing.
p (λ) = l (λ) φ (λ) + r (λ)
where r (λ) = 0 or else it has smaller degree than φ (λ) . Then
0 = l (A) φ (A) x + r (A) x
It follows that r (A) x = 0. If r (λ) 6= 0 this would be a contradiction to the definition
of φ (λ). Therefore, φ (λ) divides p (λ) .
6. Let A ∈ L (V, V ) where V is a finite dimensional vector space with field of scalars F.
Let p (λ) be the minimal polynomial and suppose φ (λ) is an irreducible polynomial
with the property that φ (A) x = 0 for some specific x 6= 0. Show that φ (λ) must
divide p (λ) . Hint: First write
p (λ) = φ (λ) g (λ) + r (λ)
where r (λ) is either 0 or has degree smaller than the degree of φ (λ). If r (λ) = 0 you
are done. Suppose it is not 0. Let η (λ) be the monic polynomial of smallest degree
with the property that η (A) x = 0. Now use the Euclidean algorithm to divide φ (λ)
by η (λ) . Contradict the irreducibility of φ (λ) .
Say φ (A) x = 0 where x 6= 0. Consider all monic polynomials φ (λ) such that φ (A) x =
0. One such is the minimal polynomial p (λ). This condition is less severe than saying
that φ (A) = 0. Therefore, there are more polynomials considered. If you take the one
of least degree, η its degree must be no larger than the degree of p (λ) . You can do
the usual thing.
p (λ) = g (λ) η (λ) + r (λ)
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
108
Exercises
where r (λ) = 0 or else it has smaller degree than η (λ) . Then
0 = g (A) η (A) x + r (A) x
It follows that r (A) x = 0. If r (λ) 6= 0 this would be a contradiction to the definition
of η (λ). Therefore, η (λ) divides p (λ) . Also
φ (λ) = η (λ) l (λ) + r (λ)
where the degree of r (λ) is less than the degree of η (λ) or else r (λ) = 0. If r (λ) 6= 0,
then r (A) x = 0 and it would have degree too small. Hence r (λ) = 0 and so η (λ)
must divide φ (λ) . But this requires that the two must be scalar multiples of each
other. Hence φ (λ) must also divide p (λ).
7. Suppose A is a linear transformation and let the characteristic polynomial be
det (λI − A) =
q
Y
φj (λ)nj
j=1
where the φj (λ) are irreducible. Explain using Corollary 8.3.11 why the irreducible
factors of the minimal polynomial are φj (λ) and why the minimal polynomial is of
the form
q
Y
r
φj (λ) j
j=1
where rj ≤ nj . You can use the Cayley Hamilton theorem if you like.
This follows right away from the Cayley Hamilton theorem and that corollary. Since
the minimal polynomial divides the characteristic polynomial, it must be of the form
just mentioned.
8. Let
1 0
A= 0 0
0 1
Find the minimal polynomial for A.
0
−1
0
You know that this satisfies a polynomial of degree no more than 3. Consider
2
3
1 0 0
1 0 0
1 0 0
1 0 0
0 1 0 , 0 0 −1 , 0 0 −1 , 0 0 −1 These are
0 0 1
0 1 0
0 1 0
0 1 0
1 0 0
1 0 0
1 0
0
1 0 0
0 1 0 , 0 0 −1 , 0 −1 0 , 0 0 1
0 0 1
0 1 0
0 0 −1
0 −1 0
Thus it is desired to find a linear combination of these which equals 0 and out of all
of them to pick the one using as few of them as possible starting with the left side.
An easy way to do this is to use the row reduced echelon from. Write them as column
vectors and then do row operations.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
109
Exercises
1
0
0
0
1
0
0
0
1
1
1
1
1 1
1
1
1
0
0
0
0 0
0
0
0
0
0
0
0 0
0
0
0
0
0
0
0 0
0
0
0
0 −1 0 , 1 0 −1 0 , row echelon form: 0
−1 0
1
0 −1 0
1
0
0
0
0
0 0
0
0
0
1
0 −1
0 1
0 −1
0
0 −1 0
1 0 −1 0
0
0
1
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
1
−1
1
0
0
0
0
0
0
Then it follows that A3 = I − A + A2 and so the minimal polynomial is,
λ3 − λ2 + λ − 1
Does it work?
1
0
0
3
0 0
0 −1 −
1 0
0 0
1 0
0 −1 − 0 1
1 0
0 0
1
0
0
2
0
−1 +
0
0
0 0 0
0 = 0 0 0
1
0 0 0
1
0
0
0
0
1
9. SupposeA is an n × n matrix
and let v be a vector. Consider the A cyclic set of
vectors v, Av, · · · , Am−1 v where this is an independent set of vectors but Am v is
a linear combination of the preceding vectors in the list. Show how to obtain a monic
polynomial of smallest degree, m, φv (λ) such that
φv (A) v = 0
Now let {w1 , · · · , wn } be a basis and let φ (λ) be the least common multiple of the
φwk (λ) . Explain why this must be the minimal polynomial of A. Give a reasonably
easy algorithm for computing φv (λ).
You could simply form the cyclic set of vectors described, make them the columns of
a matrix and then do the row reduced echelon form to reveal linear relations. This is
how you can compute φv (λ). Then, having found this, for each vector in a basis, and
letting φ (λ) be the least common multiple, it follows that φ (A) v = 0 for each v in a
basis and so φ (A) = 0. Also, if η (A) = 0, then η (λ) must be divisible by each of the
φv (λ) for v in the given basis. Here is why.
η (λ) = l (λ) φv (λ) + r (λ) , degree of r (λ) < degree of φv (λ) or r (λ) = 0
Then if r (λ) 6= 0, you could contradict the fact that φv (λ) has smallest degree out of
all those φ (λ) for which φ (A) v = 0. Therefore, η (λ) is divisible by each φv (λ) for
each v in the given basis. The polynomial of smallest such degree is by definition the
least common multiple.
10. Here is a matrix.
−7 −1
−21 −3
70
10
−1
−3
10
Using the process of Problem 9 find the minimal polynomial of this matrix. It turns
out the characteristic polynomial is λ3 .
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
110
Exercises
1 0 0
1
−7 −1 −1
1
0 0 0
1
0 1 0 0 −21 −3 −3 0 0 0 0 0
0 0 1
0
70
10 10
0
0 0 0
0
1
−7
0
1 −7 0
1 0 0
0 −21 0 , 0 −21 0 , row echelon form: 0 1 0 It fol0
70
0
0 70 0
0 0 0
2
lows that φe1 (λ) , λ = 0 + 0λ. The same sort of thing will happen with e2 , e3 .
Therefore, the minimal polynomial is just λ2 . Note that
2
−7 −1 −1
0 0 0
−21 −3 −3 = 0 0 0
70
10 10
0 0 0
11. Find the minimal polynomial for
1 2 3
A= 2 1 4
−3 2 1
by the above technique. Is what you found also the characteristic polynomial?
First here
1
0
0
1 1
0 2
0 −3
is the string of vectors which comes from e1
1
−4
−26
2 −8 −24
−3
−2
−6
−4 −26
1 0 0 −14
−8 −24 , row echelon form: 0 1 0
0
−2 −6
0 0 1
3
φe1 (λ) : −14 + 3λ2 = λ3
The minimal polynomial is then λ3 −3λ2 +14. We know it can’t have any larger degree
and so this must be it.
3
2
1 2 3
1 2 3
2 1 4 −3 2 1 4
−3 2 1
−3 2 1
0
1 2 3
0 0
+14 2 1 4 = 0 0
−3 2 1
0 0
0
0
0
This is also the characteristic equation. Note that no computation involving determinants was required.
12. Let A be an n × n matrix with field of scalars C. Letting λ be an eigenvalue, show
the dimension of the eigenspace equals the number of Jordan blocks in the Jordan
canonical form which are associated with λ. Recall the eigenspace is ker (λI − A) .
Let the blocks associated with λ in the Jordan form be
J1 (λ)
J2 (λ)
..
.
Jr (λ)
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
111
Exercises
Thus this is an upper triangular matrix which has λ down the main diagonal. If you
take away λI from it, you get a matrix of the form
N1 (λ)
N2 (λ)
..
.
Nr (λ)
where
0
1
0
Ni (λ) =
0
0
0
..
.
..
.
1
0
How many columns of zeros will there be? Exactly r where r is the number of those
blocks. You get one at the beginning of each block. Hence there will be exactly r free
variables and so the dimension of the eigenspace will equal to the number of blocks.
13. For any n × n matrix, why is the dimension of the eigenspace always less than or
equal to the algebraic multiplicity of the eigenvalue as a root of the characteristic
equation? Hint: Note the algebraic multiplicity is the size of the appropriate block
in the Jordan form.
This is obvious from the above problem. The dimension of the eigenspace equals the
number of blocks which must be no larger than the length of the diagonal.
14. Give an example of two nilpotent
minimal polynomial if possible.
2
0 0 0
0 1 0 0
0 0 0
0 0 0 0
0 0 0 0 = 0 0 0
0 0 0
0 0 0 0
2
0 1 0 0
0 0 0
0 0 0 0
0 0 0
0 0 0 1 = 0 0 0
0 0 0 0
0 0 0
These are two which work.
matrices which are not similar but have the same
0
0
,
0
0
0
0
0
0
15. Use the existence of the Jordan canonical form for a linear transformation whose
minimal polynomial factors completely to give a proof of the Cayley Hamilton theorem
which is valid for any field of scalars. Hint: First assume the minimal polynomial
factors completely into linear factors. If this does not happen, consider a splitting field
of the minimal polynomial. Then consider the minimal polynomial with respect to
this larger field. How will the two minimal polynomials be related? Show the minimal
polynomial always divides the characteristic polynomial.
First suppose the minimal polynomial factors. Then the linear transformation has a
Jordan form.
J (λ1 )
..
J =
.
J (λr )
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
112
Exercises
where each block is as described in the description of the Jordan form. Say J (λi )
is ri × ri in size. Then the characteristic polynomial is the determinant of λI − J a
matrix of the form
(λ − λ1 ) I + N (λ1 )
..
.
(λ − λr ) I + N (λr )
where the blocks N (λi ) have some −1’s on the super diagonal and zeros elsewhere.
Thus the characteristic polynomial is the determinant of this matrix which is
q (λ) =
r
Y
i=1
Then q (J) is given by
r
Y
i=1
The matrix J − λi I is of the form
J (λ1 − λi )
..
.
ri
(λ − λi )
ri
(J − λi I)
J (0)
..
.
J (λr − λi )
r
where the block J (0) is nilpotent with J (0) i = 0. It follows that (J − λi I)
form
r
J (λ1 − λi ) i
..
.
0
..
.
ri
J (λr − λi )
Qr
ri
Therefore, the product i=1 (J − λi I) = 0 from block multiplication.
ri
is of the
Now in the general case where the minimal polynomial pF (λ) does not necessarily
factor, consider a splitting field for the minimal polynomial. The new minimal polynomial divides the old one. Therefore, the new minimal polynomial with respect to
this splitting field G can be completely factored. Hence the Jordan form exists with
respect to this new splitting field and new minimal polynomial pG (λ) . Then from the
above, it follows that pG (J) = 0. However, pG (λ) divides pF (λ) and so it is also the
case that pF (J) = 0. Since J is a matrix of T with respect to a suitable basis, it
follows that pF (T ) = 0 also, where T is the original linear transformation.
16. Here is a matrix. Find its Jordan canonical form by directly finding the eigenvectors
and generalized eigenvectors based on these to find a basis which will yield the Jordan
form. The eigenvalues are 1 and 2.
−3 −2 5 3
−1 0 1 2
−4 −3 6 4
−1 −1 1 3
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
113
Exercises
Why is
−3
−1
−4
−1
it typically impossible to find the Jordan canonical form?
−2 5 3
4
1
0 1 2
1
, eigenvectors: ↔ 1, 0 ↔ 2
3
1
−3 6 4
−1 1 3
1
0
Now it is necessary to find generalized eigenvectors.
−3 −2 5 3
1 0 0 0
x1
4
−1 0 1 2 0 1 0 0 x2 1
−4 −3 6 4 − 0 0 1 0 x3 = 3
x4
1
−1 −1 1 3
0 0 0 1
4tˆ4 − 4
−4 −2 5 3
x1
4
tˆ4 + 1
−1 −1 1 2 x2 1
−4 −3 5 4 x3 = 3 , Solution is: 3tˆ4 − 2
1
x4
−1 −1 1 2
tˆ4
A generalized eigenvector is
−3 −2
−1 0
−4 −3
−1 −1
−5 −2 5
−1 −2 1
−4 −3 4
−1 −1 1
5
1
6
1
3
2
4
1
3
1 0
0 1
2
−2
0 0
4
3
0 0
x1
x2
x3 =
x4
A generalized eigenvector is
−4
1
−2
0
0
x1
x2
0
0 x3
1
x4
1
0
=
1
0
1
tˆ3
1
0
, Solution is:
tˆ3
1
0
1
0
0
1
0
So let the matrix be as follows
4
−4
1
0
1 1 0 1
3 −2 1 0 ,
1
0
0
1
0
1
0
1
4 −4 1
1 1 0
3 −2 1
1 0 0
4 −4 1
1 1 0
S=
3 −2 1
1 0 0
0
1
0
1
0
1
0
1
Now to get the Jordan form you take S −1 AS, but of course we know the Jordan form
at this point. There will be two blocks, one for 1 and one for 2. Nevertheless, here it
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
114
Exercises
is
4
1
3
1
−4
1
−2
0
−1
0
−3
−1
1
0 −4
1
−1
1
0
=
0
0
1
0
1
0
−2
0
−3
−1
1
1
0
0
5
1
6
1
0
0
2
0
3
2
4
3
4 −4 1
1 1 0
3 −2 1
1 0 0
0
0
1
2
0
1
0
1
17. People like to consider the solutions of first order linear systems of equations which
are of the form
x0 (t) = Ax (t)
where here A is an n × n matrix. From the theorem on the Jordan canonical form,
there exist S and S −1 such that A = SJS −1 where J is a Jordan form. Define
y (t) ≡ S −1 x (t) . Show y0 = Jy.
This is easy. You have
x0 = SJS −1 x
and so
0
S −1 x = J S −1 x , y0 = Jy
Now suppose Ψ (t) is an n × n matrix whose columns are solutions of the above
differential equation. Thus
Ψ0 = AΨ
Now let Φ be defined by SΦS −1 = Ψ. Show
Φ0 = JΦ.
You have
Φ = S −1 ΨS
and so
Φ0
=
S −1 Ψ0 S = S −1 AΨS = S −1 ASΦS −1 S
=
JΦ
18. In the above Problem show that
0
det (Ψ) = trace (A) det (Ψ)
and so
det (Ψ (t)) = Cetrace(A)t
This is called Abel’s formula and det (Ψ (t)) is called the Wronskian.
It is easiest to consider Φ0 = JΦ and verify that det (Φ (t)) = Cetrace(J)t . This will do
it because J and A are similar and so are Φ and Ψ.
0
det (Φ) =
n
X
det Φi
i=1
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
115
Exercises
where Φi has the ith row differentiated and the other rows of Φ left unchanged. Thus,
letting φi denote the ith row of Φ, you have that the j th entry of φ0i is given by
X
φ0ij (t) =
Jik Φkj = λi Φij + ai Φ(i+1)j
k
Thus
φ0i (t) = λi φi (t) + ai φi+1 (t)
where ai either is 0 or 1. Then since the determinant of a matrix having two equal
rows equals 0, it follows that
0
(det (Φ)) =
n
X
λi det Φ = trace (J) det (Φ)
i=1
and so
det (Φ) = Cetrace(J)t
19. Let A be an n × n matrix and let J be its Jordan canonical form. Recall J is a block
diagonal matrix having blocks Jk (λ) down the diagonal. Each of these blocks is of
the form
λ 1
0
.
λ ..
Jk (λ) =
..
. 1
0
λ
Now for ε > 0 given, let the diagonal matrix Dε be given by
1
0
ε
Dε =
.
..
k−1
0
ε
Show that Dε−1 Jk (λ) Dε has the same form as Jk (λ) but instead of ones down the
super diagonal, there is ε down the super diagonal. That is Jk (λ) is replaced with
λ ε
0
.
λ ..
..
. ε
0
λ
Now show that for A an n × n matrix, it is similar to one which is just like the Jordan
canonical form except instead of the blocks having 1 down the super diagonal, it has
ε.
That Dε−1 Jk (λ) Dε is of the
1
0
0
0 ε−1
0
0
0
ε−2
0
0
0
right form is
λ
0
0
0
0 0
0
ε−3
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
easy to see by considering
1 0 0
1 0 0
0 ε 0
λ 1 0
0 λ 1 0 0 ε2
0 0 λ
0 0 0
the 4 × 4 case.
0
0
0
ε3
7KH6D\ORU)RXQGDWLRQ
116
Exercises
1
0
0 ε−1
=
0
0
0
0
0
0
ε−2
0
=
0
0
0
λ ε
0 λε
0 0
ε−3
0 0
λ ε 0 0
0 λ ε 0
0 0 λ ε
0 0 0 λ
Finally, the Jordan form is of the form
J1
0
0
..
.
Jm
0
ε2
λε2
0
0
0
ε3
λε3
where the blocks Jk are as just described. So multiply on the right by
Dε1
0
..
.
0
Dεm
and on the left by the inverse of this matrix. Using block mutiplication you get the
desired modification. Note that this shows that every matrix in n × n matrix having
entries in C is similar to one which is close to a diagonal matrix.
20. Let A be in L (V, V ) and suppose that Ap x 6= 0 for some x 6= 0. Show that Ap ek 6= 0
for some ek ∈ {e1 , · · · , en } , a basis for V . If you have a matrix which is nilpotent,
(Am = 0 for some m) will it always be possible to find its Jordan form? Describe how
to do it if this is the case. Hint: First explain why all the eigenvalues are 0. Then
consider the way the Jordan form for nilpotent transformations was constructed in the
above.
Yes, it is possible to compute the Jordan form. In fact, the proof of existence tells how
to do it. A nilpotent matrix has all eigenvalues equal to 0. This is because if Ax =λx,
then Ak x = λk x and so for large k, you get λk x = 0. Finding the eigenvalues is always
the difficulty.
21. Suppose A is an n × n matrix and that it has n distinct eigenvalues. How do the minimal polynomial and characteristic polynomials compare? Determine other conditions
based on the Jordan Canonical form which will cause the minimal and characteristic
polynomials to be different.
If the eigenvalues are distinct, then the two polynomials are obviously the same. More
generally, A has a Jordan form
J (λ1 )
..
.
J (λr )
where each J (λk ) is an rk × rk block diagonal matrix consisting of Jordan blocks, the
λk being the distinct eigenvalues. Corresponding to one of these J (λk ) you get the
term
(λ − λk )rk
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
117
Exercises
in the characteristic equation. If J (λk ) is of the form
λk 1
..
.
λk
.
.
. 1
λk
then J − λi I is of the form
J (λ1 − λi )
..
.
J (0)
..
.
J (λr − λi )
r
where J (0) i = 0 and no smaller exponent will work. Therefore, the characteristic
polynomial and the minimal polynomial will coincide. However, if J (λk ) is composed
of many blocks, the extreme case being when it is a diagonal matrix having λk down
the diagonal, a smaller exponsne will work and so there will be a difference between
the characteristic and minimal polynomials.
For example, consider the minimal polynomial for the matrix
0 1 0 0
0 0 0 0
N =
0 0 0 1
0 0 0 0
The characteristic polynomial is just λ4 but N 2 = 0. If you had an unbroken string
of ones down the diagonal, then the minimal polynomial would also be λ4 . Getting
difference between the two polynomials involves having lots of repeated eigenvalues
and lots of blocks for the Jordan form for each eigenvalue.
22. Suppose A is a 3 × 3 matrix and it has at least two distinct eigenvalues. Is it possible
that the minimal polynomial is different than the characteristic polynomial?
You might have distinct eigenvalues in which case
or you would have the following Jordan forms.
a 1 0
a
0 a 0 or 0
0 0 b
0
the two polynomials are the same
0
a
0
0
0
b
In the first case, the two polynomials are the same. The characteristic polynomial is
(λ − a)2 (λ − b)
In the first case
a 1
0 a
0 0
0
a
0 − 0
b
0
0
a
0
0
a
0 0
a
0
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
1
a
0
0
b
0 − 0
b
0
0 0
b 0
0 b
7KH6D\ORU)RXQGDWLRQ
118
Exercises
0
= 0
0
a−b
0
0
0
0 6= 0
0
and so the characteristic polynomial and minimal
second case however,
a 0 0
a 0 0
a
0 a 0 − 0 a 0 0
0 0 b
0 0 a
0
0
= 0
0
polynomial are the same. In the
0
a
0
0 0
0 0
0 0
0
b
0 − 0
b
0
0 0
b 0
0 b
and so the minimal polynomial is different than the characteristic polynomial
23. If A is an n × n matrix of entries from a field of scalars and if the minimal polynomial
of A splits over this field of scalars, does it follow that the characteristic polynomial
of A also splits? Explain why or why not.
If the minimal polynomial splits, then the matrix has a Jordan form with respect to
this field of scalars. Thus both the characteristic polynomial and minimal polynomial
are of the form
m
Y
r
(λ − λi ) i
i=1
Hence the characteristic polynomial also splits.
24. In proving the uniqueness of the Jordan canonical form, it was asserted that if two
n × n matrices A, B are similar, then they have the same minimal polynomial and also
that if this minimal polynomial is of the form
p (λ) =
s
Y
ri
φi (λ)
i=1
where the φi (λ) are irreducible, then ker (φi (A)ri ) has the same dimension as ker (φi (B)ri ) .
Why is this so? This was what was responsible for the blocks corresponding to an
eigenvalue being of the same size.
Since the two are similar, they come from the same linear transformation T . Therefore,
this is pretty obvious. Also, if B = S −1 AS, then
ri r
r
= ker S −1 (φi (A) i ) S
ker (φi (B) i ) = ker φi S −1 AS
r
This clearly has the same dimension as ker (φi (A) i ) .
25. Show that a given matrix is non defective (diagonalizable) if and only if the minimal
polynomial has no repeated roots.
If the matrix is non defective, then its Jordan form is
J (λ1 )
..
.
J (λr )
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
119
Exercises
where each J (λk ) is a diagonal matrix, J (λk ) of size rk × rk . Then the minimal
polynomial is
Y
(λ − λi ) .
i
Conversely, suppose the minimal polynomial is of this form. Also suppose that some
J (λi ) say J (λi ) is not diagonal. Then consider
Y
(J − λi I)
i
The ith term in the product is
J1 (λ1 − λi )
..
.
Ji (0)
..
.
Jr (λr − λi )
where for some i, Ji (0) is not the zero matrix. It follows that the above product
cannot be zero either. In fact, you could let x ∈ Fri be such that J (0) x = y 6= 0.
Then
0
0
.. ..
. .
Y
(J − λi I)
x = z , z 6= 0
. .
i
.. ..
0
0
because Ji (λi − λk ) is one to one on Fri whenever k 6= i. It follows that each of the
J (λk ) must be a diagonal matrix.
26. Describe a straight forward way to determine the minimal polynomial of an n × n
matrix using row operations. Next show that if p (λ) and p0 (λ) are relatively prime,
then p (λ) has no repeated roots. With the above problem, explain how this gives a
way to determine whether a matrix is non defective.
You take the matrix A and consider the sequence
I, A, A2 , A3 , · · · , An
There is some linear combination of these which equals 0. You just want to find the
one which involves the lowest exponents. Thus you string these out to make them
2
vectors in Fn and then make them the columns of a n2 × n matrix. Find the row
reduced echelon form of this matrix. This will allow you to identify a linear relation
which gives the minimal polynomial.
If p (λ) and p0 (λ) are relatively prime, then you would need to have
Y
p (λ) =
(λ − λi )
i
the λi being distinct. Thus the minimal polynomial would have no repeated roots and
so the matrix is non defective by the above problem.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
120
Exercises
27. In Theorem 10.3.4 show that the cyclic sets can be arranged in such a way that the
length of β xi+1 divides the length of β xi .
You make this condition a part of the induction hypothesis. Then at the end when
m−1
you extend to φ (A)
(V) ⊆ ker (φ(A)) , you note that, thanks to Lemma 10.3.2,
m−1
each of the β yi for yi ∈ ker φ (A)
has length equal to a multiple of d, the degree
m−1
m−1
of φ (λ) while the β x for x ∈ φ (A)
(V ) \ ker φ (A)
each have length equal to
d thanks to the assumption that φ (λ) is irreducible over the field of scalars and each
of these is in ker (φ (A)).
28. Show that if A is a complex n × n matrix, then
a Jordan block. Note that
λ 1 0
0
0 0 1
0 1 0 0 λ 1 0
1
0 0 λ
1 0 0
A and AT are similar. Hint: Consider
0 1
λ 0
1 0 = 1 λ
0 0
0 1
0
0
λ
Such a backwards identity will work for any size Jordan block. Therefore, J is similar
to J T because you can use a similarity matrix which is block diagonal, each block
being appropriate for producing the transpose of the corresponding Jordan block.
Thus there is a matrix S such that
J = S −1 J T S
Now also J = P −1 AP for a suitable P. Therefore,
P −1 AP = S −1 P −1 AP
and so
A =
=
T
S
−1
P S −1 P T AT P T
SP −1
−1
−1
−1
PT
SP −1
AT P T
SP −1
29. Let A be a linear transformation defined on a finite dimensional vector space V . Let
the minimal polynomial be
q
Y
m
φi (λ) i
i=1
n
o
is a basis for
be the cyclic sets such that β ivi , · · · , β ivri
1
1
i
i P P
mi
i
ker (φi (A) ). Let v = i j vj . Now let q (λ) be any polynomial and suppose that
and let
β ivi , · · · , β ivri
q (A) v = 0
Show that it follows q (A) = 0.
First suppose that V is a vector space with a cyclic basis x, Ax, · · · , An−1 x such
that An x is a linear combination of the vectors just listed. Let φ (λ) be the monic
polynomial of degree n which results from this. Thus φ (A) x = 0. Then actually,
φ (A) = 0. This is because if v is arbitrary, then
v=
n−1
X
ai Ai x.
i=0
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
121
Exercises
Then
φ (A) v =
n−1
X
ai Ai φ (A) x = 0
i=0
Now suppose that q (A) x = 0. Does it follow that q (A) = 0?
q (λ) = φ (λ) l (λ) + r (λ)
where r (λ) either has smaller degree than φ (λ) or r (λ) = 0. Suppose that r (λ) 6= 0.
Then
r (A) x = 0
and the degree of r (λ) is less than n which contradicts the linear independence of
x, Ax, · · · , An−1 x
Hence r (λ) = 0. It follows that φ (λ) divides q (λ) and so q (A) = 0.
Now consider the problem of interest. Since each span β ivi is invariant, it follows
j
P
i
i
i
that q (A) vj ∈ span β vi . Therefore, 0 = i,j q (A) vj . By independence, each vector
j
in the sum equals 0 and so from the first part of the argument q (A) restricted to
ker (φi (A)mi ) equals 0. Since the whole space is a direct sum of these, it follows that
q (A) = 0.
F.28
Exercises
10.9
1. Letting A be a complex n × n matrix, in obtaining the rational canonical form, one
obtains the Cn as a direct sum of the form
span β x1 ⊕ · · · ⊕ span β xr
where β x is an ordered cyclic set of vectors, x, Ax, · · · , Am−1 x such that Am x is in
the span of the previousvectors. Now apply the Gram Schmidt process to the ordered
basis β x1 , β x2 , · · · , β xr , the vectors in each β xi listed according to increasing power
of A, thus obtaining an ordered basis (q1 , · · · , qn ) . Letting Q be the unitary matrix
which has these vectors as columns, show that Q∗ AQ equals a matrix B which satisfies
Bij = 0 if i − j ≥ 2. Such a matrix is called an upper Hessinberg matrix and this shows
that every n × n matrix is orthogonally similar to an upper Hessinberg matrix.
You note that qk is in the span of
β x1
β x2
z
}|
{ z
}|
{
x1 , · · · , Am1 x1 , x2 , · · · , Am2 x2 , · · · , · · · xl , Axl , · · · , Aj xl
where there are k vectors in the above list. Since each span β xi is A invariant, Aqk
is in the span of
β x1
β x2
z
}|
{ z
}|
{
x1 , · · · , Am1 x1 , x2 , · · · , Am2 x2 , · · · , · · · xl , Axl , · · · , Aj xl , Aj+1 xl
and by the Gram Schmidt process, this is the same as
span (q1 , · · · , qk+1 )
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
122
Exercises
It follows that if p ≥ k + 2, then
q∗p Aqk = 0
Thus the pk th entry of Q∗ AQ equals 0 if p − k ≥ 2. That is, this matrix is upper
Hessinberg.
2. In the argument for Theorem 10.2.4 it was shown that m (A) φl (A) v = v whenever
r
r
v ∈ ker (φk (A) k ) . Show that m (A) restricted to ker (φk (A) k ) is the inverse of the
rk
linear transformation φl (A) on ker (φk (A) ) .
In this theorem you had polynomials m (λ) , n (λ) such that
rk
m (λ) φl (λ) + n (λ) φk (λ)
Thus you also have
=1
m (A) φl (A) + n (A) φk (A)rk = I
r
If x ∈ ker (φk (A) k ) , then if follows that
=0
z }| {
r
m (A) φl (A) x + n (A) φk (A) k x = x
and so m (A) is indeed the inverse of φl (A).
3. Suppose A is a linear transformation and let the characteristic polynomial be
det (λI − A) =
q
Y
nj
φj (λ)
j=1
where the φj (λ) are irreducible. Explain using Corollary 8.3.11 why the irreducible
factors of the minimal polynomial are φj (λ) and why the minimal polynomial is of
the form
q
Y
r
φj (λ) j
j=1
where rj ≤ nj . You can use the Cayley Hamilton theorem if you like.
From the Cayley Hamilton theorem, the minimal polynomial divides the characteristic
polynomial and so from the mentioned corollary, it follows that the minimal polynomial
has the form described above with rj ≤ nj .
4. Find the minimal polynomial for
1 2 3
A= 2 1 4
−3 2 1
by the above technique with the field of scalars being the rational numbers. Is what
you found also the characteristic polynomial?
1 2 3
2 1 4 , characteristic polynomial: X 3 − 3X 2 + 14
−3 2 1
It is irreducible over the rationals by the rational root theorem. Therefore, the above
must be the minimal polynomial as well.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
123
Exercises
5. Show, using the rational root theorem, the minimal polynomial for A in the above
problem is irreducible with respect to Q. Letting the field of scalars be Q find the
rational canonical form and a similarity transformation which will produce it.
To find the rational canonical form, you need to look for cyclic sets. You have a single
irreducible polynomial, λ3 − 3λ + 14 = φ (λ). Also φ (A) sends every vector to 0.
Therefore, you can simply start with a vector. Take e1 .
1 2
1
0 , 2 1
0
−3 2
3
1
1 2
4 0 , 2 1
1
0
−3 2
3
1 2 3
1
2 1 4 0
−3 2 1
0
These vectors are
2
3
1
4 0
1
0
−26
1
1
−4
0 , 2 , −8 , −24
−6
0
−3
−2
The last is definitely a linear combination of the first three because the first three are
an independent set. Hence, you ought to have these as columns. Then you have
−1
1 1 −4
1 2 3
1
0 2 −8 2 1 4 0
0 −3 −2
−3 2 1
0
0
= 1
0
0 −14
0
0
1
3
1 −4
2 −8
−3 −2
and this is the rational canonical form for this matrix.
6. Find the rational canonical form for the matrix
1 2 1 −1
2 3 0 2
1 3 2 4
1 2 1 2
This was just pulled out of the air. To find the rational canonical form, look for cyclic
sets.
2
1 0 0 0
1
1 2 1 −1
1
1 2 1 −1
1
0 1 0 0 0 2 3 0 2 0 2 3 0 2 0
0 0 1 0 0 , 1 3 2 4 0 , 1 3 2 4 0
0 0 0 1
0
1 2 1 2
0
1 2 1 2
0
1
2
1
1
2
3
3
2
3
1 −1
1
0
0 2
2 4 0
1 2
0
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
1
2
,
1
1
2
3
3
2
1
0
2
1
4
−1
1
0
2
4 0
2
0
7KH6D\ORU)RXQGDWLRQ
124
Exercises
These reduce to
1
0
0
0
1
1
0 2
,
0 1
0
1
1
2
1
1
5
10
13
8
30
56
93
54
5
30
10 56
,
13 , 93
8
54
181
1
0
336
, row echelon form:
0
600
343
0
181
336
,
600
343
0
1
0
0
0
0
1
0
Thus the minimal polynomial is
0 −3
0 −1
0 −11
1
8
λ4 − 8λ3 + 11λ2 + λ + 3
Then
1
0
0
0
you can find the rational
−1
1 5 30
1 2
2 3
2 10 56
1 13 93 1 3
1 8 54
1 2
canonical form from
1 −1
1 1
0 2
0 2
2 4 0 1
1 2
0 1
this. Also,
5 30
10 56
=
13 93
8 54
0
1
0
0
0
0
1
0
0
0
0
1
−3
−1
−11
8
7. Let A : Q3 → Q3 be linear. Suppose the minimal polynomial is (λ − 2) λ2 + 2λ + 7 .
Find the rational canonical form. Can you give generalizations of this rather simple
problem to other situations?
That second factor is irreducible over the rationals and so you would need the following
for the rational canonical form.
2 0 0
0 0 −7
0 1 −2
8. Find the rational canonical form with respect to the field of scalars equal to Q for the
matrix
0 0 1
A = 1 0 −1
0 1 1
Observe that this particular matrix is already a companion matrix of λ3 − λ2 + λ − 1.
Then find the rational canonical form if the field of scalars equals C or Q + iQ.
First note that the minimal polynomial is λ3 − λ2 + λ − 1 and a use of the rational
root theorem shows that this factors over the rationals as (λ − 1) λ2 + 1 . Then you
need to locate cyclic sets starting with a vector in ker A2 + I . Lets find this first.
0 0
1 0
0 1
2
1
1 0
−1 + 0 1
1
0 0
0
1
0 = 0
1
1
Clearly the kernel is the vectors of the form
−s − t
s
t
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
1 1
0 0
1 1
7KH6D\ORU)RXQGDWLRQ
125
Exercises
Starting with one of these vectors, lets find cyclic sets.
−1
0
1 , 1
0
0
1
0 1
−1
0 0
0 −1 1 , 1 0
1 1
0
0 1
−1
0
1
1 , −1 , −1
0
1
0
2
1
−1
−1 1
1
0
These of course will be dependent since otherwise we don’t really have the minimal
polynomial. Thus the cycle can consist of only the first two. Thus we will have these
two as the first two columns and then we will have the eigenvector corresponding to
ker (A − I)
−1 0
1
ker 1 −1 −1
0
1
0
1
This eigenvector is just 0 . Then the rational canonical form of this matrix is
1
−1
−1 0 1
0 0
1 −1 0 1 0
0
1 1
0 1
− 21
− 21
=
1
2
1
2
− 21
1
2
1
2
1
2
1
2
1
−1 0 1
−1 1 −1 0
1
0
1 1
1
−1 0 1
−1 1 −1 0
1
0
1 1
0
0
1
0 0
1 0
0 1
0 −1
1 0
0 0
If the field of scalars is Q + iQ, then the minimal polynomial factors as
(λ
(λ − i) and so the rational canonical form is just the diagonal matrix,
− 1) (λ + i)
1 0 0
0 i 0
0 0 −i
9. Let q (λ) be a polynomial and C its companion matrix. Show the characteristic and
minimal polynomial of C are the same and both equal q (λ).
Let q (λ) = a0 + a1 λ + · · · + an−1 λn−1 + λn . Then its companion matrix is
0 ···
0
−a0
1 0 ···
−a1
C =
..
.
..
.
0
1
−an−1
Then you can show by induction that the characteristic polynomial of C is q (λ). See
Problem 43 on Page 66 It was also shown that q (C) = 0. (Cayley Hamilton theorem
for companion matrices.) Let p (λ) be the minimal polynomial. Thus it has degree no
more than the degree of q (λ) and thus must divide q (λ). Could it have smaller degree
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
126
Exercises
than q (λ)? No, it couldn’t. To see this, consider, for example a companion matrix
for a monic polynomial of degree 4 denoted as A. The following gives the sequence
A0 , A, A2 , A3 . Note that these are linearly independent. Watch the ones to see this.
0 0 0 a
0 0 a
ad
1 0 0 0
0 1 0 0 1 0 0 b 0 0 b a + bd
0 0 1 0 0 1 0 c 1 0 c b + cd
0 0 1 d
0 1 d d2 + c
0 0 0 1
0 a
ad
a d2 +
c
2
0 b a + bd
b
d
+
c
+
ad
0 c b + cd a + c d2 + c + bd
1 d d2 + c b + d d2 + c + cd
This pattern will occur for any sized companion matrix. You start with ones down the
main diagonal and then the string of ones moves down one diagonal and then another
and so forth. After n − 1 multiplications, you get a 1 in the lower left corner. Each
multiplication moves the ones into slots which had formerly been occupied with zeros.
That is why these are independent. Therefore, the degree of the minimal polynomial
is at least n and so the minimal and characteristic polynomials must coincide for any
companion matrix.
10. Use the existence of the rational canonical form to give a proof of the Cayley Hamilton
theorem valid for any field, even fields like the integers mod p for p a prime. The earlier
proof was fine for fields like Q or R where you could let λ → ∞ but it is not clear the
same result holds in general.
The linear transformation has a matrix of the following form,
M1
..
M =
.
Mq
mk
where Mk comes from ker (φk (λ)
Say the minimal polynomial is
) , φk (λ) being irreducible over the field of scalars.
p (λ) =
q
Y
φk (λ)
mk
k=1
Furthermore, Mk is of the form
C1k
..
.
Clkk
r
where Cjk is the companion matrix of η kj (λ) where η kj (λ) = φk (λ) j with rj ≤ mk . By
r
the above problem, the minimum polynomial for Cjk is just φk (λ) j , rj ≤ mk which
is also the characteristic polynomial of this matrix. The characteristic polynomial is
then
q
q Y
lk
Y
Y
q (λ) =
det (λI − Mk ) =
det λI − Cjk
i=1 j=1
k=1
=
q Y
lk
Y
η kj (λ) =
k=1 j=1
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
q Y
lk
Y
k=1 j=1
r
φk (λ) j , rj ≤ mk
7KH6D\ORU)RXQGDWLRQ
127
Exercises
It follows that
q (A) =
q Y
li
Y
φk (A)rj
i=1 j=1
k rj
We know that φk Cj
= 0.Also when you have a
B1
..
B=
.
Bq
and φ (λ) is a polynomial,
block diagonal matrix,
φ (B1 )
..
φ (B) =
.
φ (Bq )
Therefore, the above product for q (A) is the product of block diagonal matrices such
that every position on the diagonal has a zero matrix in at least one of the block
matrices in the product. This equals the zero matrix. Like this:
0 0 0
D 0 0
F 0 0
0 B 0 0 0 0 0 H 0
0 0 C
0 0 E
0 0 0
When you multiply the block matrices, it involves simply multiplying together the
blocks in the corresponding positions. Since at least one is zero, it follows the product
is the zero matrix.
11. Suppose you have two n× n matrices A, B whose entries are in a field F and suppose G
is an extension of F. For example, you could have F = Q and G = C. Suppose A and
B are similar with respect to the field G. Can it be concluded that they are similar
with respect to the field F? Hint: Use the uniqueness of the rational canonical form.
Do A, B have the same minimal polynomial over F? Denote these by pA (λ) and pB (λ)
You have
pA (A) = 0
and so pA (B) = 0 also because A and B are similar with respect to G. It follows that
pA (λ) must divide pB (λ) . Similar reasoning implies that pB (λ) must divide pA (λ).
Therefore, these two minimal polynomials having coefficients in F coincide. Otherwise,
you could write
pB (λ) = pA (λ) + r (λ)
where the degree of r (λ) is smaller than the common degree of pA (λ) and pB (λ).
Then plugging in λ = B, you get a contradiction to the definition of pB (λ) as having
the smallest degree possible. Say the common minimal polynomial over F is
q
Y
φi (λ)mi
i=1
where φi (λ) is irreducible over F. Thus, each of these φi is a polynomial with coefficients in F and hence in G, and so
φi (A) , φi (B)
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
128
Exercises
are similar over G. Letting β vi and β wj denote the A cyclic sets associated with
ker (φi (A))
mi
, ker (φi (B))
mi
respectively, it follows from Lemma 10.8.3 that the lengths of these β vi and β wi comprise exactly the same set of positive numbers. These lengths are identified with the
size of the Jordan blocks of the two similar matrices, φi (A) , φi (B). However, the
proof of this lemma exhibits a basis which yields the Jordan form, which happens to
involve only the field F. The companion matrices corresponding to two of these which
have the same length are therefore exactly the same, both being companion matrices
k
of φi (λ) for a suitable k. It follows that the rational canonical form for A, B over F
is exactly the same. Hence, by uniqueness, the two matrices are similar over F.
F.29
Exercises
11.4
1. Suppose the migration matrix for three
.5
.3
.2
locations is
0 .3
.8 0 .
.2 .7
Find a comparison for the populations in the three locations after a long time.
.5 0 .3
1 0 0
−0.5
0
0.3
.3 .8 0 − 0 1 0 = 0.3 −0.2
0
.2 .2 .7
0 0 1
0.2
0.2 −0.3
1
3
0
0
1 0 − 53 0
−2
10
9
3
− 15
0
0 , row echelon form: 0 1 − 10
0
10
1
1
3
0
0
0
0
− 10 0
5
5
It follows that the populations are
.6
t .9
1
where t is chosen such that the sum of the entries corresponds to the total amount at
the beginning.
P
2. Show that if i aij = 1, then if A = (aij ) , then the sum of the entries of Av equals
the sum of the entries of v. Thus it does not matter whether aij ≥ 0 for this to be so.
P P
P P
P
i
j aij vj =
j
i aij vj =
j vj
3. If A satisfies the conditions of the above problem, can it be concluded that limn→∞ An
exists?
Not necessarily. Consider the following matrix.
4
9
−3 −8
4 4
9
−311 −936
=
etc. The eigenvalues of this matrix are
−3 −8
312
937
1, −5. Thus there is no way that the limit can exist. Just do the powers of the matrix
to the eigenvector which corresponds to −5.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
129
Exercises
4. Give
−1.
1
0
0
an example of a non regular Markov matrix which has an eigenvalue equal to
0
0
1
0
1 , eigenvalues: 1, −1
0
5. Show that when a Markov matrix is non defective, all of the above theory can be proved
very easily. In particular, prove the theorem about the existence of limn→∞ An if the
eigenvalues are either 1 or have absolute value less than 1.
It suffices to show that limn→∞ An v exists for each v ∈ Rn . This is because you could
then assert that limn→∞ eTi An ej exists. But this is just the ij th entry of An . Let the
eigen pairs be
(1, v1 ) , (λi , vi ) , i = 2, · · · , n
and by assumption, these vi form a basis. Therefore, if v is any vector, there exist
unique scalars ci such that
X
v=
ci vi
i
It follows that
Am v = c1 v1 +
n
X
j=2
ci λm vi → c1 v1
Thus this is really easy.
6. Find a formula for An where
5
2
5
A=
7
2
7
2
− 21
0
− 21
− 21
0
0
1
2
0
−1
−4
− 25
−2
Does limn→∞ An exist? Note that all the rows sum to 1. Hint: This matrix is similar
to a diagonal matrix. The eigenvalues are 1, −1, 12 , 12 .
5
− 21 0 −1
2
5
0
0 −4
7
1
− 2 12 − 52
2
7
− 21 0 −2
2
−1 0 0 0
1 1 1 0
−1
0 0 1
1
3 2 1 0 0
0 0
1 0 −2
2
1
=
5
2
0 0 1 0 1
1
1
−1
0 1
4
2 1 1 0
0 0 0 12
− 41 − 41 1 − 21
Now it follows that
1 1 1 0
3 2 1 0
An =
2 5 1 1
4
2 1 1 0
1
n
1
2n − (−1) + 1
2n
2n − 3 (−1)n + 1 2n
2
2
=
1n − 2 (−1)n + 1 1n
2
2
n
1
1
2n − 2 (−1) + 1
2n
n
(−1)
0
0
0
−1
−1
−1
−1
0
−1
0
1
0
1
1
2n
0
0 1
−1
1
1
1
0
−
−
2n
4
4
n
2
0
(−1) − 2n + 1
0 3 (−1)n − 24n + 1
1
2 (−1)n − 23n + 1
2n
n
0 2 (−1) − 22n + 1
0
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
0
0
1
0
0 1
0 −2
0 1
1 − 12
7KH6D\ORU)RXQGDWLRQ
130
Exercises
7. Find a formula for An where
2
4
A=
5
2
3
− 21
0
− 21
− 21
−1
−4
−2
−2
1
2
1
1
1
2
Note that the rows sum to 1 in this matrix also. Hint: This matrix is not similar
to a diagonal matrix but you can find the Jordan form and consider this in order to
obtain a formula for this product. The eigenvalues are 1, −1, 21 , 12 .
2 − 21 12 −1
4
0
1 −4
5
1
− 2 1 −2
2
3 − 21 12 −2
− 97 1 − 61 79
−1 0 0 0
1 0
0
−1
− 23 1 − 1 14 0 1 0 0 1 −1
0
1
9
3
9
=
− 13 1 − 1 4 0 0 1 1 0 −6 −14 20
9
6
9
2
0 0 0 12
1 0
−3
2
− 16
1 − 61 79
9
Now consider the Jordan block on the bottom.
It when raising this to the nth power, you get
=
=
1
2 X
n
2n−k
k
0
k=0
1
0
2n
+n
1
0
n
1 21−n 2
n
2n
1
0
n
2
0
1
2n−k
1
2n−1
0
1
2n−1
0
0
0
1
0
k
0 1
0 0
It follows that An is given by the formula
n
(−1)
0 0
0
1 0
0
−1
− 79 1 − 61 97
− 23 1 − 1 14
1 −1
0
1 0
0
0
1
9
3
9
13
1
4
1
1−n
−
1
−
0
0
2
n
0
−6
−14
20
2n
9
6
9
1
7
1
1
0
−3
2
− 16
1
−
0
0
0
2n
9
6
9
n
n
7
1 1−n
7
1
1 1−n
7
16
1 1−n
−
2
n
−
(−1)
+
1
−
1
n
n+1
9×2n
6
9
2n
22
9 (−1) − 9×2n − 3 2
n
n
23
32
2 1−n
1−n
14 n − 1 21−n n − 23 (−1) + 1 2n − 1
2
n
n+1
3
9
2
9 (−1) − 9×2n − 3 2
9×2
4 n − 1 21−n n − 13 (−1)n + 1 1n − 1 1 21−n n + 1n 13 (−1)n − 22 n − 1 21−n n + 1
9×2
6
9
2
2
2
9
9×2
3
n
n
7
1 1−n
1
1 1−n
16
16
1 1−n
n − 16
n
n+1
9×2n − 6 2
9 (−1) + 1
2n − 1
22
9 (−1) − 9×2n − 3 2
8. Find limn→∞ An if it exists for the matrix
1
− 12
2
1
−1
2
2
A=
1
1
2
3
2
2
3
2
− 21
− 21
3
2
3
2
0
0
0
1
The eigenvalues are 12 , 1, 1, 1.
The matrix equals
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
131
Exercises
1
−1 1 −1 0
−1 1
0
0
0
2 −1 1 0 0
4 −3 1 1
0
0
1
2
0
0
0
0
1
0
0
1
1
0
0 −1
0
1
0 1
1 1
1 0
2 −1
Hence the matrix raised to the nth power is
−1 1 −1 0
1 0 0 0
1
−1 1
0 1n 0 0 1
0
0
2
2 −1 1 0 0 0 1 0 −1
4 −3 1 1
0
0 0 0 1
you get
−1 1 −1 0
1 0 0 0
1
−1 1
0 0 0 0 1
0
0
2 −1 1 0 0 0 1 0 −1
4 −3 1 1
0 0 0 1
0
0 −1 −1 0
−1 0 −1 0
=
1
1
2 0
3
3
3 1
0
1
1
2
0
1
1
2
1
1
0
−1
1
1
0
−1
0
0
0
1
0
0
and so, in the limit,
0
1
0
0
0
1
9. Given an example of a matrix A which has eigenvalues which are either equal to 1,−1,
or have absolute value strictly less than 1 but which has the property that limn→∞ An
does not exist.
1 0 0
An easy example is 0 0 1 , eigenvalues: 1, −1
0 1 0
n
1 0 0
1 0 0
0 0 1 alternates between the identity and 0 0 1 .
0 1 0
0 1 0
10. If A is an n × n matrix such that all the eigenvalues have absolute value less than 1,
show limn→∞ An = 0.
The Jordan form is
J (λ1 )
..
.
J (λr )
m
where each |λr | < 1. Then as shown above, J (λr )
→ 0.
11. Find an example of a 3 × 3 matrix A such that limn→∞ An does not exist but
limr→∞ A5r does exist.
An easy example is the following.
i(2π/5)
e
0
0
A=
0
ei(2π/5)
0
i(2π/5)
0
0
e
i(2π/5)
5
e
0
0
= I.
A5 =
0
ei(2π/5)
0
i(2π/5)
0
0
e
Therefore, limr→∞ A5r exists. However, limr→∞ Ar cannot exist because it will just
yield diagonal matrices which have various fifth roots of 1 on the diagonal.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
132
Exercises
12. If A is a Markov matrix and B is similar to A, does it follow that B is also a Markov
matrix?
No. Start with a Markov matrix
1/2 1/3
1/2 2/3
Now take a similarity transformation of it.
1
−1 −2
1 2
1/2 1/3
1 2
=
3 4
1/2 2/3
3 4
1
This is certainly not a Markov matrix.
−1
5
3
13. P
In Theorem 11.1.3
P suppose everything is unchanged except that you assume either
a
≤
1
or
ij
j
i aij ≤ 1. Would the same conclusion be valid? What if you don’t
insist that each aij ≥ 0? Would the conclusion hold in this case?
You can’t conclude that the limit exists if the sign of aij is not restricted. For example,
4
9
−3 −8
has an eigenvalue of −5Pand so the limit of powers of this matrix cannot exist. What
about the case where i aij ≤ 1 instead of equal to 1? In this case, you do still
get the conclusion of this theorem. The condition was only used to get the entries
of An bounded independent of n and this can be accomplished just as well with the
inequality.
14. Let V be an n dimensional vector space and let x ∈ V and x 6= 0. Consider β x ≡
x,Ax, · · · ,Am−1 x where
Am x ∈ span x,Ax, · · · ,Am−1 x
and m
is the smallest such
that the above inclusion in the span takes place. Show
that x,Ax, · · · ,Am−1 x must be linearly independent. Next suppose {v1 , · · · , vn }
is a basis for V . Consider β vi as just discussed, having length mi . Thus Ami vi is a
linearly combination of vi ,Avi , · · · ,Am−1 vi for m as small as possible. Let pvi (λ) be
the monic polynomial which expresses this linear combination. Thus pvi (A) vi = 0
and the degree of pvi (λ) is as small as possible for this to take place. Show that the
minimal polynomial for A must be the monic polynomial which is the least common
multiple of these polynomials pvi (λ).
Let p (λ) be the minimal polynomial. Then there exists ki (λ) , a monic polynomial
such that
p (λ) = ki (λ) pvi (λ) + ri (λ)
where ri (λ) must be zero since otherwise, you would have ri (A) vi = 0 and this would
contradict the minimality of the degree of pvi (λ). Therefore, each of these pvi (λ)
divides p (λ) . If q (λ) is their least common multiple, then it follows that q (A) = 0
because every vector is a linear combination of these vectors vi . Therefore, p (λ)
divides q (λ). Thus
q (λ) = p (λ) k (λ)
Now each pvi (λ) divides p (λ) so it is a common multiple. It follows that k (λ) = 1
since otherwise, q (λ) wouldn’t really be the least common multiple.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
133
Exercises
15. If A is a complex Hermitian n × n matrix which has all eigenvalues nonnegative, show
that there exists a complex Hermitian matrix B such that BB = A.
You have A = U ∗ DU for D a diagonal matrix having all nonnegative entries. Then
just note that B = U ∗ D1/2 U U ∗ D1/2 U, U is unitary. Clearly you should take B =
U ∗ D1/2 U.
16. Suppose A, B are n × n real Hermitian matrices and they both have all nonnegative
eigenvalues. Show that det (A + B) ≥ det (A) + det (B).
Let P 2 = A, Q2 = B where P, Q are Hermitian and nonnegative. Then
P
A+B = P Q
Q
Now apply the Cauchy Binet formula to both sides.
X
det (A + B) =
det (Ci )2 ≥ det (P )2 + det (Q)2
i
= det P 2 + det Q2 = det (A) + det (B)
where the Ci are n × n submatrices of P Q . Note that this does not depend on
the matrices being real. You could write
P
A + B = P ∗ Q∗
Q
Then
det (A + B)
X
=
det (Ci∗ ) det (Ci ) =
i
det (Ci∗ Ci )
i
≥ det P
X
2
+ det Q
2
= det (A) + det (B)
α c∗
is an (n + 1) × (n + 1) Hermitian nonnegative matrix where
b A
α is a scalar and A is n × n. Show that α must be real, c = b, and A = A∗ , A is
nonnegative, and that if α = 0, then b = 0.
α b∗
B∗ =
= B and so you must have α is real and b = c while A = A∗ .
c A∗
Suppose α = 0 and b 6= 0. I need to show that B is not nonnegative.
17. Suppose B =
B
z
}| { 0 b∗
α
∗
α x
b A
x
∗
b x
α x∗
= αb∗ x+αx∗ b + x∗ Ax
αb + Ax
=
If for some x, x∗ Ax < 0, then let α = 0 and you see that B is not nonnegative.
Therefore, A must be nonnegative. But now you could let x = b and then pick α be
large and negative and again see that B is not nonnegative. Hence if α = 0, then
b = 0. Now consider b 6= 0 and determine the sign of α. Consider
B
β
0∗
z
}| { α b∗
β
=
b A
0
=
βα
βb∗
β2α
β
0
and so for this to be nonnegative, you have to have α > 0.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
134
Exercises
18. ↑If A is an n × n complex Hermitian and nonnegative matrix, show that there exists
an upper triangular matrix B such that B ∗ B = A. Hint: Prove this by induction. It
is obviously true if n = 1. Now if you have an (n + 1) × (n
+ 21) Hermitian
nonnegative
α
αb∗
matrix, then from the above problem, it is of the form
, α real.
αb A
From the above problem, a generic (n + 1) × (n + 1) Hermitian nonnegative matrix is
of the form just claimed where A is nonnegative and Hermitian and α is real. In case
α = 0, there is nothing to prove. You would just use
0 0∗
0 0∗
0 B∗
0 B
where B ∗ B = A. Otherwise, you consider
α 0∗
α
b B∗
0
b∗
B
19. ↑ Suppose A is a nonnegative Hermitian matrix which is partitioned as
A11 A12
A=
A21 A22
where A11 , A22 are square matrices. Show that det (A) ≤ det (A11 ) det (A22 ). Hint:
Use the above problem to factor A getting
∗
B11 0∗
B11 B12
A=
∗
∗
B12
B22
0
B22
∗
∗
∗
Next argue that A11 = B11
B11 , A22 = B12
B12 + B22
B22 . Use the Cauchy Binet theo∗
∗
∗
rem to argue that det (A22 ) = det (B12
B12 + B22
B22 ) ≥ det (B22
B22 ) . Then explain
why
det (A)
∗
∗
= det (B11
) det (B22
) det (B11 ) det (B22 )
∗
∗
= det (B11
B11 ) det (B22
B22 )
∗
∗
∗
The Hint tells pretty much how to do it. A11 = B11
B11 , A22 = B12
B12 +B22
B22 follows
from block multiplication. Why is the determinant of a block triangular
matrix
equal
B11 B12
to the product of the determinants of the blocks? Consider
. If the
0
B22
rank of B22 is not maximal, then its determinant is zero and also the determinant of
the matrix is zero. Assume then that it has full rank. Then suitable row operations
can be applied to obtain that the determinant of this matrix equals the determinant
of
B11
0
0
B22
This determinant is clearly equal to the product of the determinants of B11 and B22 .
∗
∗
∗
The inequality det (B12
B12 + B22
B22 ) ≥ det (B22
B22 ) follows from the Cauchy Binet
theorem since there are more nonnegative terms in the sum for the determinant on the
left than for the term on the right. Note that these determinants are all real because
the matrices are Hermitian. It follows that
det (A)
∗
∗
= det (B11
B11 ) det (B22
B22 )
∗
∗
≤ det (A11 ) det (B12
B12 + B22
B22 ) = det (A11 ) det (A22 )
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
135
Exercises
20. ↑ Prove the inequality of Hadamard. If A is a Hermitian matrix which is nonnegative,
then
Y
det (A) ≤
Aii
i
This follows from induction. It is clearly true if n = 1. Now let A be n×n, nonnegative
and Hermitian and assume the theorem is true for all nonnegative matrices which are
of smaller size. Then
B a
A=
a∗ α
It follows that B and α are both nonnegative. Then from the above,
det (A) ≤ det (B) det (α)
By induction, this implies
det (A) ≤
F.30
Y
Aii
i
Exercises
12.7
1. Find the best solution to the system
x + 2y = 6
2x − y = 5
3x + 2y = 0
T
1 2
1 2
14 6
2 −1 2 −1 =
6 9
3 2
3 2
T
1 2
6
14 6
x
Solve
= 2 −1 5
6 9
y
3 2
0
17 14 6
x
16
15
=
, Solution is:
1
6 9
y
7
45
2. Find an orthonormal basis for R3 , {w1 , w2 , w3 } given that w1 is a multiple of the
vector (1, 1, 2).
0
1
1
A basis consists of 1 , 1 , 0
2
0
0
1 0 1
1 1 0
2 0 0
√ √
√ √
√
√
1√
1
1
1
5 6 25 5
6
6
6 √6 − 30
6 √
6 √ 6√
√
√
√
1
1
1
= 16 √6
5 √
6
0√ 0
6 √
6 5 6 − 30 √5 6
1
1
1
2
6
−
5
6
−
5
5
0
0
3
15
5
5
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
136
Exercises
Then one example of an orthonormal basis extending the given vector is
√ √ 2√
1√
1
5 6
− 30
6 √6
5 5
√
√
1 6 , 1 5 6 ,
0√
6√
6 √ √
1
1
1
−
6
−
5
6
5 5
3
15
3. Suppose A = AT is a symmetric real n × n matrix which has all positive eigenvalues.
Define
(x, y) ≡ (Ax, y) .
Show this is an inner product on Rn . What does the Cauchy Schwarz inequality say
in this case?
It satisfies all the axioms of an inner product obviously. The only ones which are not
obvious are whether (Ax, x) ≥ 0 and so forth. But A = U ∗ DU where D is diagonal
and U is orthogonal. Therefore,
2
(Ax, x) = (U ∗ DU x, x) = (DU x,U x) ≥ µ |U x| = µ |x|
2
where µ is the smallest eigenvalue which is assumed positive. Therefore, also (Ax, x) =
0 if and only if |x| = 0. The Cauchy Schwarz inequality says
|(Ax, y)| ≤ (Ax, x)1/2 (Ay, y)1/2
4. Let
||x||∞ ≡ max {|xj | : j = 1, 2, · · · , n} .
T
Show this is a norm on Cn . Here x = x1 · · · xn
. Show
1/2
||x||∞ ≤ |x| ≡ (x, x)
where the above is the usual inner product on Cn .
The axioms of a norm are all obvious except for the triangle inequality. However,
kx + yk∞ = max {|xi + yi | , i ≤ n} ≤ kxk∞ + kyk∞
so even this one is all right. Also for some i
kxk∞
1/2
2
= |xi | = |xi |
≤
5. Let
||x||1 ≡
Show this is a norm on Cn . Here x =
n
X
j=1
!
i
|xi |
2
"1/2
= |x| .
|xj | .
···
x1
X
xn
||x||1 ≥ |x| ≡ (x, x)
1/2
T
. Show
where the above is the usual inner product on Cn .
It is obvious that k·k1 is a norm. What of the inequality? Is
!
"1/2
X
X
2
|xi | ≥
|xi |
?
i
i
Of course. Just square both sides and the left has mixed terms which are not present
on the right.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
137
Exercises
6. Show that if ||·|| is any norm on any vector space, then
|||x|| − ||y||| ≤ ||x − y|| .
kxk = kx − y + yk ≤ kx − yk + kyk and so
kxk − kyk ≤ kx − yk
Now repeat the argument with the x, y switched. Finally, |kxk − kyk| equals either
kxk − kyk or kyk − kxk either way, you get the inequality.
7. Relax the assumptions in the axioms for the inner product. Change the axiom about
(x, x) ≥ 0 and equals 0 if and only if x = 0 to simply read (x, x) ≥ 0. Show the Cauchy
Schwarz inequality still holds in the following form.
|(x, y)| ≤ (x, x)
1/2
1/2
(y, y)
.
The proof given included this case.
8. Let H be an inner product space and let {uk }nk=1 be an orthonormal basis for H.
Show
n
X
(x, y) =
(x, uk ) (y, uk ).
k=1
You have x =
P
j
(x, uj ) uj and y =
(x, y) =
=
X
P
j
(y, uj ) uj . Therefore,
(x, uj ) uj ,
j
j
X
X
(y, uk ) uk
(x, uj ) (y, uk ) (uj , uk ) =
j,k
n
X
(x, uk ) (y, uk )
k=1
9. Let the vector space V consist of real polynomials of degree no larger than 3. Thus a
typical vector is a polynomial of the form
a + bx + cx2 + dx3 .
For p, q ∈ V define the inner product,
(p, q) ≡
Z
1
p (x) q (x) dx.
0
Show this is indeed an inner product.
Then state
the Cauchy Schwarz inequality in
terms of this inner product. Show 1, x, x2 , x3 is a basis for V . Finally, find an
orthonormal basis for V. This is an example of some orthonormal polynomials.
p1 (x) = 1
p2 (x) =
√
x−(1/2)
1/2 =
3 (2x − 1)
1 2
x− 2 )
0 (
√
√
x2 −(x2 , 3(2x−1)) 3(2x−1)−(x2 ,1)1
x2 −x+ 16
√
√
√ √
=
1
2
2
2
2
kx −(x , 3(2x−1)) 3(2x−1)−(x ,1)1k
kx − 6 3 3(2x−1)− 13 k
x−(x,p1 )p1
kx−(x,p1 )p1 k
=
R
1
p3 (x) =
√
= 6 5 x2 − x + 16
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
138
Exercises
√
√
√
√
x3 −(x3 ,6 5(x2 −x+ 16 ))6 5(x2 −x+ 16 )−(x3 , 3(2x−1)) 3(2x−1)− 14
√
√
√
√
kx3 −(x3 ,6 5(x2 −x+ 16 ))6 5(x2 −x+ 16 )−(x3 , 3(2x−1)) 3(2x−1)− 14 k
√ √
√ √
√ √
√ √
1
1
3
3
x3 − 20
56 5(x2 −x+ 16 )− 20
3 3(2x−1)− 41
x3 − 20
56 5(x2 −x+ 16 )− 20
3 3(2x−1)− 14
√
√ √
√ √
=
1
1
1
1
3
3
2
2800
kx − 20 56 5(x −x+ 6 )− 20 3 3(2x−1)− 4 k
2800
p4 (x) =
=
√
= 20 7 x3 − 32 x2 + 35 x −
1
20
It is clear the vectors given form a basis. The orthonormal basis
√
√
√
1
3
2
1, 3 (2x − 1) , 6 5 x − x +
, 20 7 x3 − x2 +
6
2
is
3
1
x−
5
20
10. Let Pn denote the polynomials of degree no larger than n − 1 which are defined on an
interval [a, b] . Let {x1 , · · · , xn } be n distinct points in [a, b] . Now define for p, q ∈ Pn ,
(p, q) ≡
n
X
p (xj ) q (xj )
j=1
Show this yields an inner product on Pn . Hint: Most of the axioms are obvious. The
one which says (p, p) = 0 if and only if p = 0 is the only interesting one. To verify this
one, note that a nonzero polynomial of degree no more than n − 1 has at most n − 1
zeros.
If (p, p) = 0 then p equals zero at n distinct points and so it must be the zero polynomial
because it has degree at most n − 1 and so it can have no more than n − 1 roots.
11. Let C ([0, 1]) denote the vector space of continuous real valued functions defined on
[0, 1]. Let the inner product be given as
Z 1
(f, g) ≡
f (x) g (x) dx
0
Show this is an inner product. Also let V be the subspace described in Problem 9.
Using the result of this problem, find the vector in V which is closest to x4 .
It equals
√
√
x4 , 1 1 + x4 , 3 (2x − 1)
3 (2x − 1)
√
√
1
1
+ x4 , 6 5 x2 − x +
6 5 x2 − x +
6
6
√
√
3
3
1
3
3
1
+ x4 , 20 7 x3 − x2 + x −
20 7 x3 − x2 + x −
2
5
20
2
5
20
Now it is just a matter of working these out.
√
√
R1 √
2
x4 , 3 (2x − 1) = 0 x4 3 (2x − 1) = 15
3
√
√
R 1 4 √
1
2
4
2
2
x , 6 5 x − x + 6 = 0 x 6 5 x − x + 16 = 35
5
√
√
√
R
1
1
1
1
x4 , 20 7 x3 − 32 x2 + 53 x − 20
= 0 x4 20 7 x3 − 32 x2 + 35 x − 20
= 70
7
Therefore, the closest point is
1
2√ √
2√ √
1
+
3 3 (2x − 1) +
56 5 x2 − x +
+
5 15
35
6
√
1√
3
3
1
+
720 7 x3 − x2 + x −
70
2
5
20
2
1
9
= 2x3 − x2 + x −
7
7
70
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
139
Exercises
12. A regular Sturm Liouville problem involves the differential equation, for an unknown function of x which is denoted here by y,
0
(p (x) y 0 ) + (λq (x) + r (x)) y = 0, x ∈ [a, b]
and it is assumed that p (t) , q (t) > 0 for any t ∈ [a, b] and also there are boundary
conditions,
C1 y (a) + C2 y 0 (a) = 0
C3 y (b) + C4 y 0 (b) = 0
where
C12 + C22 > 0, and C32 + C42 > 0.
There is an immense theory connected to these important problems. The constant, λ
is called an eigenvalue. Show that if y is a solution to the above problem corresponding
to λ = λ1 and if z is a solution corresponding to λ = λ2 6= λ1 , then
Z
b
q (x) y (x) z (x) dx = 0.
(6.37)
a
and this defines an inner product. Hint: Do something like this:
0
(p (x) y 0 ) z + (λ1 q (x) + r (x)) yz
0
(p (x) z 0 ) y + (λ2 q (x) + r (x)) zy
= 0,
= 0.
Now subtract and either use integration by parts or show
0
0
0
(p (x) y 0 ) z − (p (x) z 0 ) y = ((p (x) y 0 ) z − (p (x) z 0 ) y)
and then integrate. Use the boundary conditions to show that y 0 (a) z (a)−z 0 (a) y (a) =
0 and y 0 (b) z (b) − z 0 (b) y (b) = 0. The formula, 6.37 is called an orthogonality relation.
It turns out there are typically infinitely many eigenvalues and it is interesting to write
given functions as an infinite series of these “eigenfunctions”.
Let y go with λ and z go with µ.
0
z (p (x) y 0 ) + (λq (x) + r (x)) yz
0
y (p (x) z 0 ) + (µq (x) + r (x)) zy
Subtract.
0
=
=
0
0
0
z (p (x) y 0 ) − y (p (x) z 0 ) + (λ − µ) q (x) yz = 0
Now integrate from a to b. First note that
0
0
z (p (x) y 0 ) − y (p (x) z 0 ) =
d
(p (x) y 0 z − p (x) z 0 y)
dx
and so what you get is
p (b) y 0 (b) z (b) − p (b) z 0 (b) y (b) − (p (a) y 0 (a) z (a) − p (a) z 0 (a) y (a))
+ (λ − µ)
Z
b
q (x) y (x) z (x) dx = 0
a
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
140
Exercises
Look at the stuff on the top line. From the assumptions on the boundary conditions,
C1 y (a) + C2 y 0 (a)
0
C1 z (a) + C2 z (a)
= 0
= 0
and so
y (a) z 0 (a) − y 0 (a) z (a) = 0
Similarly,
y (b) z 0 (b) − y 0 (b) z (b) = 0
Hence, that stuff on the top line equals zero and so the orthogonality condition holds.
13. Consider the continuous functions defined on [0, π] , C ([0, π]) . Show
Z π
(f, g) ≡
f gdx
0
nq
o∞
2
is an inner product on this vector space. Show the functions
sin
(nx)
are
π
n=1
an orthonormal set. What does
of the vector space
q the dimension
qthis mean about
2
2
C ([0, π])? Now let VN = span
π sin (x) , · · · ,
π sin (N x) . For f ∈ C ([0, π]) find
a formula for the vector in VN which is closest to f with respect to the norm determined
from the above inner product. This is called the N th partial sum of the Fourier series
of f . An important problem is to determine whether and in what way this Fourier
series converges to the function f . The norm which comes from this inner product is
sometimes called the mean square norm.
Note that sin (nx) is a solution to the boundary value problem
y 00 + n2 y = 0, y (0) = y (π) = 0
It is obvious that sin (nx) solves this boundary value problem. Those boundary conditions in the above problem are implied by these. In fact,
1 sin 0 + (0) cos (0) =
1 sin π + (0) cos π
=
0
0
Then it follows that
Z
π
sin (nx) sin (mx) dx = 0 unless n = m.
0
Now also
Rπ
0
sin2 (nx) dx = 21 π and so
n√
√2
π
sin (nx)
o
are orthonormal.
14. Consider the subspace V ≡ ker (A) where
1 4 −1 −1
2 1 2
3
A=
4 9 0
1
5 6 3
4
Find an orthonormal basis for V. Hint: You might first find a basis and then use the
Gram Schmidt procedure.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
141
Exercises
1
2
4
5
4
1
9
6
−1 −1
x
y
2
3
0
1 z
3
4
w
9
− 7 tˆ3
0
0
4 tˆ3
= , Solution is: 7
0
tˆ3
0
0
This is pretty easy. The space has dimension 1. Therefore, an orthonormal basis is
just
√
9
146
− 146
−9
√
4 2 146
1
= 73 √
√
7
146
16 + 49 + 81 7 146
0
0
15. The Gram Schmidt process starts with a basis for a subspace {v1 , · · · , vn } and produces an orthonormal basis for the same subspace {u1 , · · · , un } such that
span (v1 , · · · , vk ) = span (u1 , · · · , uk )
for each k. Show that in the case of Rm the QR factorization does the same thing.
More specifically, if
A = v1 · · · vn
and if
A = QR ≡
q1
···
qn
R
then the vectors {q1 , · · · , qn } is an orthonormal set of vectors and for each k,
span (q1 , · · · , qk ) = span (v1 , · · · , vk )
This follows from the way we multiply matrices and the definition of an orthogonal
matrix.
16. Verify the parallelogram identify for any inner product space,
|x + y|2 + |x − y|2 = 2 |x|2 + 2 |y|2 .
Why is it called the parallelogram identity?
2
2
|x + y| + |x − y| = (x + y, x + y) + (x − y, x − y)
= |x|2 + |y|2 + 2 (x, y) + |x|2 + |y|2 − 2 (x, y) .
17. ↑Let H be an inner product space and let K ⊆ H be a nonempty convex subset. This
means that if k1 , k2 ∈ K, then the line segment consisting of points of the form
tk1 + (1 − t) k2 for t ∈ [0, 1]
is also contained in K. Suppose for each x ∈ H, there exists P x defined to be a point
of K closest to x. Show that P x is unique so that P actually is a map. Hint: Suppose
z1 and z2 both work as closest points. Consider the midpoint, (z1 + z2 ) /2 and use the
parallelogram identity of Problem 16 in an auspicious manner.
Suppose there are two closest points to x say k1 , k2 both work. Then
K and
2 2
x − k1 + k2 = x − k1 + x − k2 2
2
2
2
2 6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
k1 +k2
2
is also in
7KH6D\ORU)RXQGDWLRQ
142
Exercises
=
=
2
2
k2 − k1 2
+ 2 x − k1 + 2 x − k2 − 2
2
2
2
2 2
k2 − k1 + d2
− 2 where d is the distance between x and ki .
18. ↑In the situation of Problem 17 suppose K is a closed convex subset and that H
is complete. This means every Cauchy sequence converges. Recall from calculus a
sequence {kn } is a Cauchy sequence if for every ε > 0 there exists Nε such that
whenever m, n > Nε , it follows |km − kn | < ε. Let {kn } be a sequence of points of K
such that
lim |x − kn | = inf {|x − k| : k ∈ K}
n→∞
This is called a minimizing sequence. Show there exists a unique k ∈ K such that
limn→∞ |kn − k| and that k = P x. That is, there exists a well defined projection map
onto the convex subset of H. Hint: Use the parallelogram identity in an auspicious
manner to show {kn } is a Cauchy sequence which must therefore converge. Since K
is closed it follows this will converge to something in K which is the desired vector.
Let x be given and let {kn } be a minimizing squence.
|x − kn | → λ ≡ inf {|x − y| : y ∈ K}
Then
2
x − kn 2
x − km 2
kn − km 2 x − kn
x − km +
+
= 2
+2
2
2
2 2 2 Now it follows that
kn − km 2
2
=
≤
1
2
|x − kn | +
2
1
2
|x − kn | +
2
2
1
kn + km 2
|km − x| − x −
2
2
1
2
|km − x| − λ2
2
The right side converges to 0 as n, m → ∞ and so {kn } is a Cauchy sequence and
since the space is given to be complete, it follows that this converges to some k. Since
K is closed, it follows that k ∈ K. Now
|x − k| = lim |x − kn | = λ
n→∞
and so there exists a closest point.
19. ↑Let H be an inner product space which is also complete and let P denote the projection map onto a convex closed subset, K. Show this projection map is characterized
by the inequality
Re (k − P x, x − P x) ≤ 0
for all k ∈ K. That is, a point z ∈ K equals P x if and only if the above variational
inequality holds. This is what that inequality is called. This is because k is allowed
to vary and the inequality continues to hold for all k ∈ K.
Consider for t ∈ [0, 1] the following.
|y − (x + t (w − x))|
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
2
7KH6D\ORU)RXQGDWLRQ
143
Exercises
where w ∈ K and x ∈ K. It equals
2
2
f (t) = |y − x| + t2 |w − x| − 2t Re hy − x, w − xi
Suppose x is the point of K which is closest to y. Then f 0 (0) ≥ 0. However, f 0 (0) =
−2 Re hy − x, w − xi . Therefore, if x is closest to y,
Re hy − x, w − xi ≤ 0.
Next suppose this condition holds. Then you have
2
2
2
|y − (x + t (w − x))| ≥ |y − x| + t2 |w − x| ≥ |y − x|
2
By convexity of K, a generic point of K is of the form x + t (w − x) for w ∈ K. Hence
x is the closest point.
20. ↑Using Problem 19 and Problems 17 - 18 show the projection map, P onto a closed
convex subset is Lipschitz continuous with Lipschitz constant 1. That is
|P x − P y| ≤ |x − y|
hP x − P y,y − P yi ≤
hP y − P x,x − P xi ≤
0
0
Thus
hP x − P y,x − P xi ≥ 0
Hence
hP x − P y,x − P xi − hP x − P y,y − P yi ≥ 0
and so
hP x − P y,x−y − (P x − P y)i ≥ 0
2
|x−y| |P x − P y| ≥ hP x − P y,P x − P yi = |P x − P y|
21. Give an example of two vectors in R4 x, y and a subspace V such that x · y = 0 but
P x·P y 6= 0 where P denotes the projection map which sends x to its closest point on
V.
Try this. V is the span of e1 and e2 and x = e3 + e1 , y = e4 + e1 .
P x = (e3 + e1 , e1 ) e1 + (e3 + e1 , e2 ) e2 = e1
P y = (e4 + e1 , e1 ) e1 + (e4 + e1 , e2 ) e2 = e1
P x·P y = 1
22. Suppose you are given the data, (1, 2) , (2, 4) , (3, 8) , (0, 0) . Find the linear regression
line using the formulas derived above. Then graph the given data along with your
regression line.
You draw the graphs. You want to solve
a+b=2
2a + b = 4
3a + b = 8
b=0
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
144
Exercises
Of course there is no solution. The least squares solution is to solve
T
T
1 1
1 1
1 1
2
2 1 2 1 a
2 1 4
=
3 1 3 1 b
3 1 8
0 1
0 1
0 1
0
13 14 6
a
34
5
=
, Solution is:
Then the best line is
6 4
b
14
− 52
y=
13
2
x−
5
5
23. Generalize the least squares procedure to the situation in which data is given and you
desire to fit it with an expression of the form y = af (x) + bg (x) + c where the problem
would be to find a, b and c in order to minimize the error. Could this be generalized
to higher dimensions? How about more functions?
Say you had ordered pairs given which you wanted the curve to go through. Say these
are (xi , yi ) for i ≤ n. Then you would be finding least squares solutions for a, b, c to
yi = af (xi ) + bg (xi ) + c, i = 1, · · · , n
You would simply obtain a least squares solutions to this. The principles are the same
for higher dimensions.
24. Let A ∈ L (X, Y ) where X and Y are finite dimensional vector spaces with the dimension of X equal to n. Define rank (A) ≡ dim (A (X)) and nullity(A) ≡ dim (ker (A)) .
r
Show that nullity(A) + rank (A) = dim (X) . Hint: Let {xi }i=1 be a basis for ker (A)
r
n−r
n−r
and let {xi }i=1 ∪ {yi }i=1 be a basis for X. Then show that {Ayi }i=1 is linearly
independent and spans AX.
m
r
Let {Ayi }i=1 be a basis for AX and let {xi }i=1 be a basis for ker (A).
Then {y1 , · · · , ym , x1 , · · · , xr } is linearly independent. In fact it spans X because if
z ∈ X, Then
m
X
Az =
ci Ayi
i=1
Pm
r
and so z − i=1 ci yi ∈ ker (X) and so it is in the span of the {xi }i=1 . Therefore,
z is in the span of the given vectors and so this list of vectors is a basis. Hence the
dimension of X is equal to m + r = rank (A) + dim (ker (A)).
25. Let A be an m×n matrix. Show the column rank of A equals the column rank of A∗ A.
Next verify column rank of A∗ A is no larger than column rank of A∗ . Next justify the
following inequality to conclude the column rank of A equals the column rank of A∗ .
rank (A) = rank (A∗ A) ≤ rank (A∗ ) ≤
= rank (AA∗ ) ≤ rank (A) .
r
r
Hint: Start with an orthonormal basis, {Axj }j=1 of A (Fn ) and verify {A∗ Axj }j=1
is a basis for A∗ A (Fn ) .
Pr
Say you have i=1 ci A∗ Axi = 0. Then for any appropriate y,
!
!
r
r
X
X
∗
0=
ci A Axi , y =
ci Axi , Ay
i=1
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
i=1
7KH6D\ORU)RXQGDWLRQ
145
Exercises
In particular, you could have y =
P
i ci xi .
r
X
Then you would get
ci Axi = 0
i=1
and so each ci = 0. Now if you have x ∈ Fn , Then Ax is a linear combination of the
Axj and so A∗ Ax is a linear combination of the A∗ Axi which shows that these vectors
are a basis for A∗ AFn . It follows that the rank of A and the rank of A∗ A are the
same. Similarly, the rank of A∗ and AA∗ are the same. Hence the above inequalities
follow right away.
26. Let A be a real m × n matrix and let A = QR be the QR factorization with Q
orthogonal and R upper triangular. Show that there exists a solution x to the equation
RT Rx = RT QT b
and that this solution is also a least squares solution defined above such that AT Ax =
AT b.
There exists a soluton to this last equation. Hence you have a solution to
RT QT QRx = RT QT b
But this is just the equations
RT Rx = RT QT b
Of course this kind of thing is very easy for a computer to solve.
F.31
Exercises
12.9
T
T
T
1. Here are three vectors in R4 : (1, 2, 0, 3) , (2, 1, −3, 2) , (0, 0, 1, 2) . Find the three
dimensional volume of the parallelepiped determined by these three vectors.
(1, 2, 0, 3) · (1, 2, 0, 3) = 14
(1, 2, 0, 3) · (2, 1, −3, 2) = 10
(1, 2, 0, 3) · (0, 0, 1, 2) = 6
(2, 1, −3, 2) · (2, 1, −3, 2) = 18
(2, 1, −3, 2) · (0, 0, 1, 2) = 1
(0, 0, 1, 2) · (0, 0, 1, 2) = 5
volume is
√
218
14 10 6
det 10 18 1 = 218
6 1 5
T
T
2. Here are two vectors in R4 : (1, 2, 0, 3) , (2, 1, −3, 2) . Find the volume of the parallelepiped determined by these two vectors.
(1, 2, 0, 3) · (1, 2, 0, 3) = 14
(1, 2, 0, 3) · (2, 1, −3, 2) = 10
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
146
Exercises
(2, 1, −3, 2) · (2, 1, −3, 2) = 18
det
volume is
14 10
10 18
T
T
= 152
√
152
T
3. Here are three vectors in R2 : (1, 2) , (2, 1) , (0, 1) . Find the three dimensional
volume of the parallelepiped determined by these three vectors. Recall that from the
above theorem, this should equal 0.
It does equal 0.
4. Find the equation of the plane through the three points (1, 2, 3) , (2, −3, 1) , (1, 1, 7) .
5. Let T map a vector space V to itself. Explain why T is one to one if and only if T is
onto. It is in the text, but do it again in your own words.
This is because if T is one to one, it takes a basis to a basis. If T is onto, then it takes
a basis to a spanning set which must be independent, since otherwise, there would be
a linearly independent set of vectors which would also span and yet have fewer vectors
than a basis.
6. ↑Let all matrices be complex with complex field of scalars and let A be an n×n matrix
and B a m × m matrix while X will be an n × m matrix. The problem is to consider
solutions to Sylvester’s equation. Solve the following equation for X
AX − XB = C
where C is an arbitrary n × m matrix. If q (λ) is a polynomial, show first that if
AX − XB = 0, then q (A) X − Xq (B) = 0. Next define the linear map T which maps
the n × m matrices to the n × m matrices as follows.
T X ≡ AX − XB
Show that the only solution to T X = 0 is X = 0 if and only if σ (A) ∩ σ (B) = ∅.
Conclude that there exists a unique solution for each C to Sylvester’s equation if and
only if σ (A) ∩ σ (B) = ∅.
q (A) X − Xq (B) = q (A) X = 0
Now explain why q (A)−1 exists if and only if σ (A) ∩ σ (B) = ∅.
Consider the first part. Multiply by An−1 . Then An−1 AX − An−1 XB = 0. Since
AX = XB, you can make multiple switches and write this as
An X − XB n = 0
Now it follows that the desired result holds for a polynomial. Consider T which maps
the vector space of n × m matrices to itself. Let q (λ) be the characteristic polynomial
of B as suggested and suppose T X = 0. Then
q (A) X − Xq (B) = 0 = q (A) X.
Qm
Let the characteristic polynomial for B be i=1 (λ − λi ) . In case there are no shared
eigenvalues, it follows that
m
Y
q (A) =
(A − λi I)
i=1
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
147
Exercises
is invertible. Hence X = 0. Therefore, T is one to one and it follows that it is onto. If
there is a shared eigenvalue, then the matrix q (A) will no longer be invertible. Hence
there will exist nonzero solutions X to q (A) X = 0. It follows that in this case T is
not one to one and so it is not onto either.
7. Compare Definition 12.8.2 with the Binet Cauchy theorem, Theorem 3.3.14. What is
the geometric meaning of the Binet Cauchy theorem in this context?
Letting U be the matrix having ui as the ith column the above definition states that
1/2
the volume is just det U T U
and by the Cauchy Binet theorem, in the interesting
case that U has more rows than columns, this equals
!
X
2
det (UI )
I
"1/2
where I is a choice of m rows which are selected from U so that UI is the square matrix
which results from these rows. The geometric meaning of the above expression is this.
det (UI ) gives the m dimensional volume of the parallelepiped which is obtained from
a projection onto the m dimensional subspace of Rn determined by setting the other
n − m variables equal to 0. Thus the volume is the square root of the sum of the
squares of the areas of these projections.
F.32
Exercises
13.12
∗
∗
1. Show (A∗ ) = A and (AB) = B ∗ A∗ .
∗
∗
(A∗ ) x, y = (x, A∗ y) = (Ax, y) . Since y is arbitrary, (A∗ ) x = Ax.
∗
(AB) x, y = (x, ABy) = (A∗ x, By) = (B ∗ A∗ x, y)
∗
Since y is arbitrary, this shows (AB) x = B ∗ A∗ x and since x is arbitrary, this shows
the formula.
2. Prove Corollary 13.3.9.
This is just like the proof of the theorem.
3. Show that if A is an n × n matrix which has an inverse then A+ = A−1 .
You have AA−1 A = A, A−1 AA−1 = A−1 , and AA−1 = I which is Hermitian and
A−1 A = I which is also Hermitian.
4. Using the singular value decomposition, show that for any square matrix A, it follows
that A∗ A is unitarily similar to AA∗ .
You have A = U ∗ ΣV where Σ is the singular value matrix and U, V are unitary of the
right size. Therefore, A∗ A = V ∗ ΣU U ∗ ΣV = V ∗ Σ2 V . Similarly, AA∗ = U ∗ Σ2 U. Then
Σ2 = U AA∗ U ∗ and so
A∗ A = V ∗ U AA∗ U ∗ V
Since these matrices are all square, this does it. U ∗ V is unitary.
5. Let A, B be a m × n matrices. Define an inner product on the set of m × n matrices
by
(A, B)F ≡ trace (AB ∗ ) .
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
148
Exercises
Show this is an inner product
P satisfying all the inner product axioms. Recall for M an
n × n matrix, trace (M ) ≡ ni=1 Mii . The resulting norm, ||·||F is called the Frobenius
norm and it can be used to measure the distance between two matrices.
This was done earlier.
6. Let A be an m × n matrix. Show
||A||2F ≡ (A, A)F =
X
σ 2j
j
where the σ j are the singular values of A.
2
kAkF = trace (AA∗ ) = trace (U ∗ ΣV V ∗ Σ∗ U ) = trace (ΣΣ∗ ) = sum of the squares of
the singular values.
7. If A is a general n × n matrix having possibly repeated eigenvalues, show there is a
sequence {Ak } of n × n matrices having distinct eigenvalues which has the property
that the ij th entry of Ak converges to the ij th entry of A for all ij. Hint: Use Schur’s
theorem.
A = U ∗ T U where T is upper triangular and U is unitary. Change the diagonal entries
of T slightly so that the resulting upper triangular matrix Tk has all distinct diagonal
entries and Tk → T in the sense that the ij th entry of Tk converges to the ij th entry
of T . Then let Ak = U ∗ Tk U. It follows that Ak → A in the sense that corresponding
entries converge.
8. Prove the Cayley Hamilton theorem as follows. First suppose A has a basis of eigenn
vectors {vk }k=1 , Avk = λk vk . Let p (λ) be the characteristic polynomial. Show
p (A) vk = p (λk ) vk = 0. Then since {vk } is a basis, it follows p (A) x = 0 for all
x and so p (A) = 0. Next in the general case, use Problem 7 to obtain a sequence {Ak }
of matrices whose entries converge to the entries of A such that Ak has n distinct
eigenvalues and therefore by Theorem 7.1.7 Ak has a basis of eigenvectors. Therefore, from the first part and for pk (λ) the characteristic polynomial for Ak , it follows
pk (Ak ) = 0. Now explain why and the sense in which
lim pk (Ak ) = p (A) .
k→∞
First say A has a basis of eigenvectors {v1 , · · · , vn } . Ak vjP= λkj vj . Then it follows
n
that p (A) vk = p (λk ) vk . Hence if x is any vector, let x = k=1 xk vk and it follows
that
!
n
n
X
X
p (A) x = p (A)
xk vk =
xk p (A) vk
k=1
=
n
X
xk p (λk ) vk =
k=1
k=1
n
X
xk 0vk = 0
k=1
Hence p (A) = 0. Now drop the assumption that A is nondefective. From the above,
there exists a sequence Ak which is non defective which converges to A and also
pk (λ) → p (λ) uniformly on compact sets because these characteristic polynomials
are defined in terms of determinants of the corresponding matrix. See the above
construction of the Ak . It is probably easiest to use the Frobinius norm for the last
part.
kpk (Ak ) − p (A)kF ≤ kpk (Ak ) − p (Ak )kF + kp (Ak ) − p (A)kF
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
149
Exercises
The first term converges to 0 because the convergence of Ak to A implies all entries
of Ak lie in a compact set. The second term converges to 0 also because the entries of
Ak converge to the corresponding entries of A.
9. Prove that Theorem 13.4.6 and Corollary 13.4.7 can be strengthened so that the
condition
on
the Ak is necessary as well as sufficient. Hint: Consider vectors of the
x
form
where x ∈ Fk .
0
Say the matrix is positive definite. Then you could just look at
x
T
x
0 M (A)
= xT M (A)k x
0
and this needs to be positive. Hence those principle minors have all positive determinants.
10. Show directly that if A is an n × n matrix and A = A∗ (A is Hermitian) then all the
eigenvalues and eigenvectors are real and that eigenvectors associated with distinct
eigenvalues are orthogonal, (their inner product is zero).
¯ (x, x) so λ = λ.
¯ Now if x is an eigenvector,
λ (x, x) = (Ax, x) = (x,Ax) = λ
A¯
x =Ax = λ¯
x and so x + x
¯ is also an eigenvector. Hence, you can assume that
the eigenvectors are real. If x
¯ is pure imaginary, you could simply multiply by i and
get an eigenvector which is real.
11. Let v1 , · · · , vn be an orthonormal basis for Fn . Let Q be a matrix whose ith column
is vi . Show
Q∗ Q = QQ∗ = I.
This follows from how we multiply matrices.
12. Show that an n × n matrix Q is unitary if and only if it preserves distances. This
means |Qv| = |v| . This was done in the text but you should try to do it for yourself.
13. Suppose {v1 , · · · , vn } and {w1 , · · · , wn } are two orthonormal bases for Fn and suppose Q is an n × n matrix satisfying Qvi = wi . Then show Q is unitary. If |v| = 1,
show there is a unitary transformation which maps v to e1 .
This is easy because you show it preserves distances.
14. Finish the proof of Theorem 13.6.5.
15. Let A be a Hermitian matrix so A = A∗ and suppose all eigenvalues of A are larger
than δ 2 . Show
2
(Av, v) ≥ δ 2 |v|
Where here, the inner product is
(v, u) ≡
n
X
vj uj .
j=1
Since it is Hermitian, there exists unitary U such that U ∗ AU = D. Then
(Ax, x) = (U DU ∗ x, x) = (DU ∗ x,U ∗ x) ≥ δ 2 |U ∗ x|2 = δ 2 |x|2
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
150
Exercises
16. Suppose A + A∗ has all negative eigenvalues. Then show that the eigenvalues of A
have all negative real parts.
You have
0 >
=
((A + A∗ ) x, x) = (Ax, x) + (A∗ x, x)
(Ax, x) + (Ax, x)
Now let Ax = λx. Then you get
2
2
¯ |x| = Re (λ) |x|
0 > λ |x| + λ
2
17. The discrete Fourier transform maps Cn → Cn as follows.
n−1
1 X −i 2π
e n jk xj .
F (x) = z where zk = √
n j=0
Show that F −1 exists and is given by the formula
n−1
1 X i 2π jk
F −1 (z) = x where xj = √
e n zk
n j=0
Here is one way to approach this problem. Note z = U x
2π
2π
2π
e−i n 0·0
e−i n 1·0
e−i n 2·0
2π
2π
2π
e−i n 1·1
e−i n 2·1
e−i n 0·1
2π
2π
1 e−i 2π
n 0·2
e−i n 1·2
e−i n 2·2
U=√
n
..
..
..
.
.
.
2π
2π
2π
e−i n 0·(n−1) e−i n 1·(n−1) e−i n 2·(n−1)
where
2π
···
···
···
e−i n (n−1)·0
2π
e−i n (n−1)·1
2π
e−i n (n−1)·2
..
.
···
e−i n (n−1)·(n−1)
2π
Now argue U is unitary and use this to establish the result. To show this verify
each row has length 1 and the inner product of two different rows gives 0. Now
2π
2π
Ukj = e−i n jk and so (U ∗ )kj = ei n jk .
Take the inner product of two rows.
n−1
X
2π
2π
e−i n jk e−i n jl
=
n−1
X
2π
2π
e−i n jk ei n jl =
j=0
j=0
=
n−1
X
2π
e−i n j(l−k)
j=0
1−e
−i 2π
n n(l−k)
2π
1 − e−i n (l−k)
= 0 if l 6= k
Pn−1
2π
because e−i n n(l−k) = 1. What if l = k? then the sum reduces to j=0 √1n √1n = 1
and so this matrix is unitary. Therefore, its inverse is just the transpose conjugate
which yields the other formula.
18. Let f be a periodic function having period 2π. The Fourier series of f is an expression
of the form
∞
n
X
X
ikx
ck e ≡ lim
ck eikx
n→∞
k=−∞
k=−n
and the idea is to find ck such that the above sequence converges in some way to f . If
f (x) =
∞
X
ck eikx
k=−∞
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
151
Exercises
and you formally multiply both sides by e−imx and then integrate from 0 to 2π,
interchanging the integral with the sum without any concern for whether this makes
sense, show it is reasonable from this to expect
Z 2π
1
cm =
f (x) e−imx dx.
2π 0
Now suppose you only know f (x) at equally spaced points 2πj/n for j = 0, 1, · · · , n.
Consider the Riemann sum for this integral obtained
n from using the left endpoint of
the subintervals determined from the partition 2π
j
. How does this compare with
n
j=0
the discrete Fourier transform? What happens as n → ∞ to this approximation?
Multiply by e−imx and formally integrate both sides. This leads to the formula for
cm . Now use a left sum. This would be
n−1 2π
1 X
2π
2π
−im 2π
j
n
f
j e
(j + 1) −
j
2π j=0
n
n
n
=
n−1 2π
2π
1X
f
j e−im n j
n j=0
n
Thus an approximation to cm comes from multiplying
above where f is the vector whose j th component is f
√1 f by the
n
√
2π
n j / n.
unitary matrix U
19. Suppose A is a real 3 × 3 orthogonal matrix (Recall this means AAT = AT A = I. )
having determinant 1. Show it must have an eigenvalue equal to 1. Note this shows
there exists a vector x 6= 0 such that Ax = x. Hint: Show first or recall that any
orthogonal matrix must preserve lengths. That is, |Ax| = |x| .
If Ax = λx, then you can take the norm of both sides and conclude that |λ| = 1. It
follows that the eigenvalues of A are eiθ , e−iθ and another one which has magnitude
1 and is real. This can only be 1 or −1. Since the determinant is given to be 1, it
follows that it is 1. Therefore, there exists an eigenvector for the eigenvalue 1.
20. Let A be a complex m × n matrix. Using the description of the Moore Penrose inverse
in terms of the singular value decomposition, show that
−1
lim (A∗ A + δI)
δ→0+
A∗ = A+
where the convergence happens in the Frobenius norm. Also verify, using the singular
value decomposition, that the inverse exists in the above formula.
Recall that the Moore Penrose inverse is
−1
σ
0
V
U∗
0
0
σ 0
where A = U
V ∗ . The matrices U, V are unitary and of the right size. First
0 0
of all, observe that there is no problem about the existence of the inverse mentioned
in the formula. This is because
((A∗ A + δI) x, x) ≥ δ |x|
2
and so it is a one to one matrix. Now consider the limit assertion. The left side equals
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
152
Exercises
−1 σ 0
σ 0
σ
V
UU
V ∗ + δI
V
0 0
0 0
0
δ→0+
2
−1 σ 0
σ 0
= lim V
V ∗ + δI
U∗
V
0 0
0 0
δ→0
lim
Now
2
σ
V
0
=
=
0
0
2
σ
V
0
2
σ
V
0
= V
= V
0
0
0
0
∗
V + δI
∗
V + δV V
+ δI
σ 2 + δI 0
0
δI
! 2
−1
σ + δI
0
=V
−1
V
∗
−1 σ
0
−1
V
σ
0
0
0
0
0
σ
0
0
0
U∗
U∗
0
0
U∗
U∗
−1 σ 0
U∗
0 0
"
σ 0
0
U∗
0 0
δ −1 I
!
−1
σ 2 + δI
σ
0
0
0
U∗
!
−1
Of course σ and σ 2 and so forth are diagonal matrices. Thus σ 2 + δI
is also a
! 2
−1
diagonal matrix having on the diagonal an expression of the form σ i + δ
σ i and
this clearly converges as δ → 0 to σ −1
.
Therefore,
the
above
converges
to
i
−1
σ
0
V
U ∗ = A+
0
0
+
21. Show that A+ = (A∗ A) A∗ . Hint: You might use the description of A+ in terms of
the singular value decomposition.
−1
σ
0
σ 0
Recall that A+ = V
U ∗ where A = U
V ∗ . The σ was the
0
0
0 0
diagonal matrix which consisting of square roots of eigenvalues of A∗ A, arranged in
decreasing order from top left toward lower right.
2
σ 0
A∗ A = V
V∗
0 0
2
+
where the σ 2 are the square roots of the eigenvalues of (A∗ A) . Thus (A∗ A) A∗ =
−2
−1
σ
0
σ 0
σ
0
∗
∗
V
V V
U =V
U ∗ = A+
0
0
0 0
0
0
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
153
Exercises
F.33
Exercises
14.7
1. Solve the system
using the
solving it
4 1
1 5
0 2
1
1 1
x
5 2 y = 2
3
2 6
z
4
1
0
1 1
x
1
7 2 y = 2
2 4
z
3
1 1
x
1
7 2 y = 2
2 4
z
3
Gauss Seidel method and the Jacobi method. Check your answer by also
using row operations.
5
1
x
1
94
7
2 y = 2 , Solution is: 94
67
4
z
3
94
3. Solve the system
using the
solving it
5 1
1 7
0 2
4
1
0
Gauss Seidel method and the Jacobi method. Check your answer by also
using row operations.
9
1
x
1
100
21
2 y = 2 , Solution is: 100
43
3
z
6
100
2. Solve the system
using the
solving it
4 1
1 7
0 2
5
1
0
Gauss Seidel method and the Jacobi method. Check your answer by also
using row operations.
5
1
x
1
118
9
2 y = 2 , Solution is: 118
42
4
z
3
59
4. If you are considering a system of the form Ax = b and A−1 does not exist, will either
the Gauss Seidel or Jacobi methods work? Explain. What does this indicate about
finding eigenvectors for a given eigenvalue?
No. These methods only work when there is a unique solution.
5. For ||x||∞ ≡ max {|xj | : j = 1, 2, · · · , n} , the parallelogram identity does not hold.
Explain.
Let x = (1, 0) , y = (0, 1) . Then
kx + yk2 + kx − yk2
2
2 kxk + 2 kyk
2
= 1+1=2
= 4
6. A norm ||·|| is said to be strictly convex if whenever ||x|| = ||y|| , x 6= y, it follows
x + y 2 < ||x|| = ||y|| .
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
154
Exercises
Show the norm |·| which comes from an inner product is strictly convex.
Let x, y be as described.
If x 6= y, then
x + y 2 x − y 2
1
1
2
2
2
+
2 2 = 2 kxk + 2 kyk = kxk
x + y 2
2
2 < kxk
7. A norm ||·|| is said to be uniformly convex if whenever ||xn || , ||yn || are equal to 1 for
all n ∈ N and limn→∞ ||xn + yn || = 2, it follows limn→∞ ||xn − yn || = 0. Show the
norm |·| coming from an inner product is always uniformly convex. Also show that
uniform convexity implies strict convexity which is defined in Problem 6.
The usual norm satisfies the parallelogram law. Thus
|xn − yn |
2
=
=
2
2
2 |xn | + 2 |yn | − |xn + yn |
4 − |xn + yn |
2
2
and the right side converges to 0. Thus |xn − yn | → 0.
8. Suppose A : Cn → Cn is a one to one and onto matrix. Define
||x|| ≡ |Ax| .
Show this is a norm.
If kxk = 0, then Ax = 0 ans since A is one to one, it follows that x = 0. Clearly
kcxk = |c| kxk. What about the triangle inequality? This is also easy.
kx + yk ≡ |A (x + y)| ≤ |Ax| + |Ay| = kxk + kyk
9. If X is a finite dimensional normed vector
space
and A, B ∈ L (X, X) such that
||B|| < ||A|| , can it be concluded that A−1 B < 1?
1 0
1/2 0
1 0
−1
Maybe not. Say A =
,B =
. Then A =
and
0 1/2
0 3/4
0 2
1
0
2
A−1 B =
so it has norm 3/2 > 1. Also kBk < kAk.
0 23
10. Let X be a vector space with a norm ||·|| and let V = span (v1 , · · · , vm ) be a finite
dimensional subspace of X such that {v1 , · · · , vm } is a basis for V. Show V is a closed
subspace of X. This means that if wn → w and each wn ∈ V, then so is w. Next show
that if w ∈
/ V,
dist (w, V ) ≡ inf {||w − v|| : v ∈ V } > 0
is a continuous function of w and
|dist (w, V ) − dist (w1 , V )| ≤ kw1 − wk
Next show that if w ∈
/ V , there exists z such that ||z|| = 1 and dist (z, V ) > 1/2.
For those who know some advanced calculus, show that if X is an infinite dimensional
vector space having norm ||·|| , then the closed unit ball in X cannot be compact. Thus
closed and bounded is never compact in an infinite dimensional normed vector space.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
155
Exercises
P
1/2
Pm
m
2
Say wn = k=1 cnk vk . Define a norm on V by kwk ≡
where w =
k=1 |ck |
Pm
c
v
.
This
gives
a
norm
on
V
which
is
equivalent
to
the
given
norm
on V .
k=1 k k
Pm
Thus if wn → w, it follows that cn → c in Cm and so w =
c
v
and
so
k=1 k k
V is closed. Let w, w0 be two points and suppose without loss of generality that
dist (w, V ) − dist (w0 , V ) ≥ 0. Let v ∈ V be such that kw0 − vk < dist (w0 , V ) + ε.
Then
|dist (w, V ) − dist (w0 , V )| =
≤
≤
=
dist (w, V ) − dist (w0 , V )
kw − vk − kw0 − vk + ε
kw − w0 k + kw0 − vk − kw0 − vk + ε
kw − w0 k + ε
Then since ε is arbitrary, this shows that |dist (w, V ) − dist (w0 , V )| ≤ kw − w0 k and
so this is continuous as claimed. Now let w ∈
/ V. Thus dist (w, V ) > 0 because if this
is not so, there wouild exist vn → w and by the first part, w ∈ V . Let v ∈ V be such
w−v
that 2 dist (w, V ) > kw − vk. Then you can consider z = kw−vk
. If v1 ∈ V,
kz − v1 k =
≥
≥
w−v
v1 kw − vk −
kw − vk
kw − vk 1
kw − v − kw − vk v1 k
2 dist (w, V )
dist (w, V )
1
=
2 dist (w, V )
2
Since v1 is arbitrary, it follows that dist (z, V ) ≥ 12 . If you have an infinite dimensional
normed linear space, the above argument shows that you can obtain an infinite sequence of vectors {xn } , each kxn k = 1 and such that dist (xn+1 , span (x1 , · · · , xn )) ≥
1/2 for every n. Therefore, there is no convergent subsequence because no subsequence
is a Cauchy sequence. Therefore, the unit ball is not sequentially compact.
11. Suppose ρ (A) < 1 for A ∈ L (V, V ) where V is a p dimensional vector space having
a norm ||·||. You can use Rp or Cp if you like. Show there exists a new norm |||·|||
such that with respect to this new norm, |||A||| < 1 where |||A||| denotes the operator
norm of A taken with respect to this new norm on V ,
|||A||| ≡ sup {|||Ax||| : |||x||| ≤ 1}
Hint: You know from Gelfand’s theorem that
||An ||
1/n
<r<1
provided n is large enough, this operator norm taken with respect to ||·||. Show there
exists 0 < λ < 1 such that
A
ρ
< 1.
λ
You can do this by arguing the eigenvalues of A/λ are the scalars µ/λ where µ ∈ σ (A).
Now let Z+ denote the nonnegative integers.
n A |||x||| ≡ sup n x
λ
n∈Z+
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
156
Exercises
First show this is actually a norm. Next explain why
n+1 A
|||Ax||| ≡ λ sup n+1 x ≤ λ |||x||| .
λ
n∈Z+
First, it is obvious that there exists λ < 1 such that
A
<1
ρ
λ
Just pick λ larger than the absolute value of all the eigenvalues which are each less
than 1. Now use the norm suggested. First of all, why is this a norm? If |||x||| = 0,
then kxk = 0 because
kxk ≤ |||x|||
It is also clear that |||cx||| = |c| |||x|||. What about the triangle inequality?
n
n n A
A A |||x + y||| ≡ sup n (x + y) ≤ sup λn x + λn y
λ
n∈Z+
n∈Z+
n n A A n y ≡ |||x||| + |||y|||
+
sup
≤ sup x
n λ λ
n∈Z
n∈Z
+
+
Now why is the operator norm with respect to this new norm less than 1?
n
n+1 A
A
|||Ax||| ≡ sup n Ax = λ sup n+1 x ≤ λ |||x|||
λ
λ
n∈Z+
n∈Z+
12. Establish a similar result to Problem 11 without using Gelfand’s theorem. Use an
argument which depends directly on the Jordan form or a modification of it.
ρ (A) < 1 and A = S −1 Jε S where Jε is in modified Jordan form having ε or 0 on the
super diagonal the diagonal entries of every block being less than 1 in absolute value.
Let λ < 1 be larger than the absolute values of all eigenvalues. Now what if you define
|||x||| ≡ kSxk∞ ? Then
|||Ax||| ≡
≤
≤
kSAxk∞ = kJε Sxk∞
max µi (Sx)i + ε (Sx)i+1 ≤
(λ + ε) kSxk∞ < kSxk∞ ≡ |||x|||
i
max {λ |(Sx)i |} + ε max {|(Sx)i |}
i
i
provided ε is chosen small enough.
13. Using Problem 11 give an easier proof of Theorem 14.6.6 without having to use Corollary 14.6.5. It would suffice to use a different norm of this problem and the contraction
mapping principle of Lemma 14.6.4.
The suggestion gives it away. You were given ρ B −1 C < 1. Now with the above
problem, there is another norm such that with respect to this other norm, B −1 C is a
contraction mapping. Therefore, it has a fixed point by the easier result on contraction
maps.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
157
Exercises
P
14. A matrix A is diagonally dominant if |aii | > j6=i |aij | . Show that the Gauss Seidel
method converges if A is diagonally dominant.
It is the eigenvalues of B −1 C which are important. Say
B −1 Cx = λx
Then you must have
det (λB − C) = 0
Illustrating what happens in the Gauss Seidel method in three dimensions, this requires
λa11 −a12 −a13
det λa21 λa22 −a23 = 0
λa31 λa32 λa33
In other words the above matrix must have a zero determinant. However, since the
original matrix was given to be diagonally dominant, this matrix cannot have a zero
determinant unless |λ| < 1.
P∞
15. Suppose f (λ) = k=0 an λn converges if |λ| < R. Show that if ρ (A) < R where A is
an n × n matrix, then
∞
X
f (A) ≡
an An
k−0
converges in L (Fn , Fn ) . Hint: Use Gelfand’s theorem and the root test.
Since the given series converges if |λ| < R, it follows that
lim sup |an λn |
1/n
<1
n→∞
and so
lim sup |an |
1/n
n→∞
|λ| < 1 if |λ| < R
Thus you have
lim sup |an |
1/n
n→∞
n 1/n
You need to look at kan A k
lim sup kan An k
≤
1
R
.
1/n
lim sup |an |
1/n
kAn k
≤
lim sup |an |
1/n
lim sup kAn k
<
1
R=1
R
=
n→∞
n→∞
n→∞
1/n
1/n
n→∞
and so the series converges.
16. Referring to Corollary 14.4.3, for λ = a + ib show
exp (λt) = eat (cos (bt) + i sin (bt)) .
Hint: Let y (t) = exp (λt) and let z (t) = e−at y (t) . Show
z 00 + b2 z = 0, z (0) = 1, z 0 (0) = ib.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
158
Exercises
Now letting z = u + iv where u, v are real valued, show
u00 + b2 u
= 0, u (0) = 1, u0 (0) = 0
v 00 + b2 v
= 0, v (0) = 0, v 0 (0) = b.
Next show u (t) = cos (bt) and v (t) = sin (bt) work in the above and that there is at
most one solution to
w00 + b2 w = 0 w (0) = α, w0 (0) = β.
Thus z (t) = cos (bt) + i sin (bt) and so y (t) = eat (cos (bt) + i sin (bt)). To show there
is at most one solution to the above problem, suppose you have two, w1 , w2 . Subtract
them. Let f = w1 − w2 . Thus
f 00 + b2 f = 0
and f is real valued. Multiply both sides by f 0 and conclude
!
2
2
d (f 0 )
2f
+b
=0
dt
2
2
Thus the expression in parenthesis is constant. Explain why this constant must equal
0.
The first claim is a routine computation. The next claim is also routine computations.
What about the uniqueness assertion? It suffices to show that if α, β = 0 then the
only solution is 0. Multiply both sides of the differential equation by w0 . Then you get
It follows that
b2 d " 2 1 d
2
(w0 ) +
w =0
2 dt
2 dt
2
(w0 ) + w2
is a constant. From the initial conditions, this constant can only be 0. The rest follows.
17. Let A ∈ L (Rn , Rn ) . Show the following power series converges in L (Rn , Rn ).
∞ k k
X
t A
k=0
k!
You might want to use Lemma 14.4.2. This is how you can define exp (tA). Next show
using arguments like those of Corollary 14.4.3
d
exp (tA) = A exp (tA)
dt
so that this is a matrix valued solution to the differential equation and initial condition
Ψ0 (t) = AΨ (t) , Ψ (0) = I.
This Ψ (t) is called a fundamental matrix for the differential equation y0 = Ay. Show
t → Ψ (t) y0 gives a solution to the initial value problem
y0 = Ay, y (0) = y0 .
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
159
Exercises
The series converges absolutely by the ratio test and the observation that Ak ≤
k
kAk . Now we need to show that you can differentiate it. Writing the difference
quotient gives
k
k
∞
∞
∞
k−1 k
k−1 k
(t
+
h)
−
t
Ak
X
X
X
1
ks (k, h)
A
s (k, h)
A
=
=
h
k!
k!
(k − 1)!
k=0
k=1
k=1
where s (k, h) is between t and t + h. Thus for each k, limh→0 s (k, h) = t. I want to
P∞ k−1 Ak
show that the limit of the series is k=1 t(k−1)!
.
∞
k−1 k
X
s (k, h)
A
k=1
=
(k − 1)!
k−1
∞
∞
s (k, h)
− tk−1 Ak
X
X
tk−1 Ak
−
=
(k − 1)!
(k − 1)!
k=1
k=1
∞
X
(k − 1) p (k, h)k−2 (s (k, h) − t) Ak
(k − 1)!
k=1
=
∞
X
p (k, h)k−2 (s (k, h) − t) Ak
(k − 2)!
k=2
where p (k, h) is also between t and t + h. Now the norm of this is dominated by
|h|
∞
k−2
k
X
|p (k, h)|
kAk
(k − 2)!
k=2
The series converges by the ratio test and so
∞
k−1 k
X
s (k, h)
A
k=1
(k − 1)!
−
∞
X
tk−1 Ak
→0
(k − 1)!
k=1
as h → 0. Hence you can differentiate the series and it gives you
∞
∞ k−1 k−1
X
X
t
A
tk−1 Ak
=A
= A exp (At)
(k − 1)!
(k − 1)!
k=1
k=1
When you plug in t = 0, you get I. Therefore, this is the solution to the given
differential equation.
18. In Problem 17 Ψ (t) is defined by the given series. Denote by exp (tσ (A)) the numbers
exp (tλ) where λ ∈ σ (A) . Show exp (tσ (A)) = σ (Ψ (t)) . This is like Lemma 14.4.7.
Letting J be the Jordan canonical form for A, explain why
Ψ (t) ≡
∞ k k
X
t A
k=0
k!
=S
∞ k k
X
t J
k=0
k!
S −1
and you note that in J k , the diagonal entries are of the form λk for λ an eigenvalue
of A. Also J = D + N where N is nilpotent and commutes with D. Argue then that
∞ k k
X
t J
k=0
k!
is an upper triangular matrix which has on the diagonal the expressions eλt where
λ ∈ σ (A) . Thus conclude
σ (Ψ (t)) ⊆ exp (tσ (A))
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
160
Exercises
Next take etλ ∈ exp (tσ (A)) and argue it must be in σ (Ψ (t)) . You can do this as
follows:
Ψ (t) − etλ I
∞ k k
∞ k k
X
X
t A
t λ
−
I
k!
k!
=
k=0
k=0
∞ k X
t
Ak − λk I
k!
k=0
∞ k k−1
X
X
t
Ak−j λj (A − λI)
k! j=1
=
=
(6.38)
k=0
Now you need to argue
∞ k k−1
X
t X
k=0
n
k!
Ak−j λj
j=1
n
converges to something in L (R , R ). To do this, use the ratio test and Lemma 14.4.2
after first using the triangle inequality. Since λ ∈ σ (A) , Ψ (t) − etλ I is not one to one
and so this establishes the other inclusion. You fill in the details. This theorem is a
special case of theorems which go by the name “spectral mapping theorem”.
P∞ (tA)k x
If λ ∈ σ (A) , then Ax = λx and so
= eλt x and so exp (tσ (A)) ⊆
k=0
k!
σ (exp (tA)) . Now consider exp (tA) . You have A = S −1 JS where J is Jordan form.
Thus it is simple to write that
exp (tA) = S −1
∞
k
X
(tJ)
k=0
k!
S
Now J is block diagonal and the blocks have a constant down the main diagonal. Say
α 1
..
.
. ..
α 1
α
When you raise such a block to the exponent k, you get an upper triangular matrix
which has αk down the diagonal. Therefore, exp (tA) is of the form
S −1 BS
where B is an upper triangular matrix which has for its diagonal entries eλt for λ ∈
σ (A). These diagonal entries are the eigenvalues of exp (tA). It follows that
σ (exp (tA)) ⊆ exp (tσ (A)) .
19. Suppose Ψ (t) ∈ L (V, W ) where V, W are finite dimensional inner product spaces and
t → Ψ (t) is continuous for t ∈ [a, b]: For every ε > 0 there there exists δ > 0 such that
if |s − t| < δ then ||Ψ (t) − Ψ (s)|| < ε. Show t → (Ψ (t) v, w) is continuous. Here it is
the inner product in W. Also define what it means for t → Ψ (t) v to be continuous
and show this is continuous. Do it all for differentiable in place of continuous. Next
show t → ||Ψ (t)|| is continuous.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
161
Exercises
|(Ψ (t) v, w) − (Ψ (s) v, w)| ≤ |Ψ (t) v − Ψ (s) v| |w|
≤ kΨ (t) − Ψ (s)k |v| |w|
and so this is continuous.
kΨ (t) v − Ψ (s) vk ≤ kΨ (t) − Ψ (s)k |v|
so t → Ψ (t) v is continuous. Differentiable works out similarly. If t → Ψ (t) is
differentiable, then those other things are too. For example,
kΨ (t + h) v − Ψ (t) v − Ψ0 (t) hvk
kΨ (t + h) − Ψ (t) − Ψ0 (t) hk |v|
o (h) |v| = o (h)
≤
=
As to continuity of t → kΨ (t)k ,
|kΨ (t)k − kΨ (s)k| ≤ kΨ (t) − Ψ (s)k
from the triangle inequality.
20. If z (t) ∈ W, a finite dimensional inner product space, what does it mean for t → z (t)
to be continuous or differentiable? If z is continuous, define
Z
b
a
as follows.
w,
Z
z (t) dt ∈ W
b
z (t) dt
a
!
≡
Z
b
(w, z (t)) dt.
a
Show that this definition is well defined and furthermore the triangle inequality,
Z
Z
b
b
z (t) dt ≤
|z (t)| dt,
a
a
and fundamental theorem of calculus,
Z t
d
z (s) ds = z (t)
dt
a
hold along with any other interesting properties of integrals which are true.
Differentiability is the same as usual.
z (t + h) − z (t) − z 0 (t) h = o (h)
If t → z (t) is continuous, then from the above problem, so is t → (z (t) , w) and so
Rb
(z (t) , w) dt makes sense. Does there exist an element I of the inner product space
a
Rb
such that (w, I) = a (w, z (t)) dt? This is so in a finite dimensional inner product
space because
Z b
w→
(w, z (t)) dt
a
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
162
Exercises
is a linear map and so it can be represented by a unique I ∈ W so that
Z b
(w, I) =
(w, z (t)) dt
a
for all w. Now what about the fundamental theorem of calculus?
!
R t+h
Z
z (s) ds
1 t+h
t
w,
≡
(w, z (s)) ds → (w, z (t))
h
h t
Since the space is finite dimensional, this is the same as saying that
R t+h
z (s) ds
= z (t) .
lim t
h→0
h
You can just apply the above to the finitely many Fourier coefficients. Therefore, the
usual Fundamental theorem of calculus holds. What about the triangle inequality?
Z
!
Z b
b
z (t) dt = sup w,
z (t) dt a
|w|=1 a
Z
Z b
b
(w, z (t)) dt ≤ sup
|(w, z (t))| dt
= sup |w|=1 a
|w|=1 a
Z b
≤
|z (t)| dt.
a
21. In the situation of Problem 19 define
Z b
Ψ (t) dt ∈ L (V, W )
a
as follows.
w,
Z
b
Ψ (t) dt (v)
a
!
≡
Z
b
(w, Ψ (t) v) dt.
a
Rb
Show this is well defined and does indeed give a Ψ (t) dt ∈ L (V, W ) . Also show the
triangle inequality
Z
Z
b
b
Ψ (t) dt ≤
||Ψ (t)|| dt
a
a
where ||·|| is the operator norm and verify the fundamental theorem of calculus holds.
Z t
0
Ψ (s) ds = Ψ (t) .
a
Also verify the usual properties of integrals continue to hold such as the fact the
integral is linear and
Z b
Z c
Z c
Ψ (t) dt +
Ψ (t) dt =
Ψ (t) dt
a
b
a
and similar things. Hint: On showing the triangle inequality, it will help if you use
the fact that
|w|W = sup |(w, v)| .
|v|≤1
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
163
Exercises
You should show this also.
That last assertion follows by the Cauchy Schwarz inequality which shows the right
side is no larger than |w| and then by taking v = w/ |w|. That the integral is well
Rb
defined follows because w → a (w, Ψ (t) v) dt is a linear mapping and so, by the
Riesz representation theorem, there exists a unique I (v) ∈ W such that (w, I (v)) =
Rb
(w, Ψ (t) v) dt. Now also, I is a linear function of v because of the fact that each
a
Rb
Ψ (t) is linear. Therefore, you can denote by a Ψ (t) dt ∈ L (V, W ) this I. Thus the
definition of the integral is well defined. The standard properties of the integral are
now fairly obvious. As for the triangle inequality, it goes the same way as in the above
problem.
Z
Z
!
Z b
b
b
Ψ (t) dt ≡ sup Ψ (t) dt (v) = sup sup w,
Ψ (t) dt (v) a
|v|≤1
|v|≤1 |w|≤1
a
a
Z
Z
b
b
= sup sup (w, Ψ (t) (v)) dt ≤
kΨ (t)k dt.
|v|≤1 |w|≤1 a
a
Now consider the fundamental theorem of calculus. For simplicity, let h → 0 + .
Z
Z
1 t+h
1 t+h
Ψ (s) ds − Ψ (t) = Ψ (s) − Ψ (t) ds
h t
h t
Z t+h
1
≤
kΨ (s) − Ψ (t)k ds
h t
which converges to 0 by continuity of Ψ.
22. Prove Gronwall’s inequality. Suppose u (t) ≥ 0, continuous, and for all t ∈ [0, T ] ,
u (t) ≤ u0 +
Z
t
Ku (s) ds.
0
where K is some nonnegative constant. Then
u (t) ≤ u0 eKt .
Rt
Hint: w (t) = 0 u (s) ds. Then using the fundamental theorem of calculus, w (t)
satisfies the following.
u (t) − Kw (t) = w0 (t) − Kw (t) ≤ u0 , w (0) = 0.
Now use the usual techniques you saw in an introductory differential equations class.
Multiply both sides of the above inequality by e−Kt and note the resulting left side is
now a total derivative. Integrate both sides from 0 to t and see what you have got. If
you have problems, look ahead in the book. This inequality is proved later in Theorem
C.4.3.
Following the hint,
" −Kt 0
we
≤ u0 e−Kt
and therefore,
w (t) e
−Kt
≤ u0
Z
t
0
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
e
−Ks
ds = u0
e−Kt
1
−
K
K
7KH6D\ORU)RXQGDWLRQ
164
Exercises
Therefore, w (t) ≤ u0
eKt
K
−
1
K
. Also, you get
u (t) ≤ u0 + Kw (t)
If K > 0, this implies
u (t) ≤ u0 + K u0
eKt
1
−
K
K
= u0 eKt
If K = 0, then the result requires no proof.
23. With Gronwall’s inequality and the integral defined in Problem 21 with its properties
listed there, prove there is at most one solution to the initial value problem
y0 = Ay, y (0) = y0 .
Hint: If there are two solutions, subtract them and call the result z. Then
z0 = Az, z (0) = 0.
It follows
z (t) = 0+
Z
t
Az (s) ds
0
and so
||z (t)|| ≤
Z
0
t
kAk ||z (s)|| ds
Now consider Gronwall’s inequality of Problem 22. Use Problem 20 as needed.
From Gronwall’s inequality,
kz (t)k ≤ 0e−kAkt = 0.
24. Suppose A is a matrix which has the property that whenever µ ∈ σ (A) , Re µ < 0.
Consider the initial value problem
y0 = Ay, y (0) = y0 .
The existence and uniqueness of a solution to this equation has been established above
in preceding problems, Problem 17 to 23. Show that in this case where the real parts
of the eigenvalues are all negative, the solution to the initial value problem satisfies
lim y (t) = 0.
t→∞
Hint: A nice way to approach this problem is to show you can reduce it to the
consideration of the initial value problem
z0 = Jε z, z (0) = z0
where Jε is the modified Jordan canonical form where instead of ones down the main
diagonal, there are ε down the main diagonal. You have y0 = S −1 Jε Sy and so you
just let z = Sy. Then z → 0 if and only if y → 0.
Then
z0 = Dz + Nε z
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
165
Exercises
where D is the diagonal matrix obtained from the eigenvalues of A and Nε is a nilpotent
matrix commuting with D which is very small provided ε is chosen very small. Now
let Ψ (t) be the solution of
Ψ0 = −DΨ, Ψ (0) = I
described earlier as
∞
X
(−1)k tk Dk
.
k!
k=0
Thus Ψ (t) commutes with D and Nε . Tell why.
The reason these commute is that D is block diagonal, each block having constant
entries down the diagonal while the same is true of Nε . The blocks associated with Nε
commute with the blocks associated with D and so D and Nε commute. Thus Ψ (t)
commutes with D and Nε .
Next argue
(Ψ (t) z)0 = Ψ (t) Nε z (t)
and integrate from 0 to t.
0
Just do the differentiation. (Ψ (t) z) =
Ψ0 (t) z (t) + Ψ (t) z0 (t) =
=
−DΨ (t) z (t) + Ψ (t) (D + Nε ) z (t)
Ψ (t) Nε z (t)
Then integrating,
Ψ (t) z (t) − z0 =
Z
t
Ψ (s) Nε z (s) ds.
0
It follows from the triangle inequality discussed above that
Z t
||Ψ (t) z (t)|| ≤ ||z0 || +
||Nε || ||Ψ (s) z (s)|| ds.
0
It follows from Gronwall’s inequality
||Ψ (t) z (t)|| ≤ ||z0 || e||Nε ||t
Now look closely at the form of Ψ (t) to get an estimate which is interesting. Explain
why
µt
e 1
0
..
Ψ (t) =
.
µn t
0
e
and now observe that if ε is chosen small enough, ||Nε || is so small that each component
of z (t) converges to 0.
It follows right away from the definition of Ψ (t) that Ψ (t) is given by the above
diagonal matrix where here the µi are negatives of the eigenvalues of A. Thus their
real parts are all bounded below by some δ > 0. It follows that
|(Ψ (t) x)i | ≥ eδt |xi |
and so
||Ψ (t) z (t)|| ≥ eδt kz (t)k
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
166
Exercises
Therefore, from the above inequality,
kz (t)k ≤ kz0 k e−δt + e−δt
Z
t
0
||z0 || e||Nε ||s ds
Of course the right side converges to 0 if you choose ε small enough that kNε k < δ.
25. Using Problem 24 show that if A is a matrix having the real parts of all eigenvalues
less than 0 then if
Ψ0 (t) = AΨ (t) , Ψ (0) = I
it follows
lim Ψ (t) = 0.
t→∞
Hint: Consider the columns of Ψ (t)?
From the above, problem, this happens. If Ψ is just defined, then the solution to
x0i = Axi (t) , xi (0) = ei is Ψ (t) ei . This follows from the fact that Ψ (t) ei satisfies the
equation and uniqueness. Thus the ith column of Ψ (t) converges to 0. In particular
the entries of Ψ (t) converge to 0. Thus kΨ (t)k → 0 also.
26. Let Ψ (t) be a fundamental matrix satisfying
Ψ0 (t) = AΨ (t) , Ψ (0) = I.
n
Show Ψ (t) = Ψ (nt) . Hint: Subtract and show the difference satisfies Φ0 = AΦ, Φ (0) =
0. Use uniqueness.
27. If the real parts of the eigenvalues of A are all negative, show that for every positive
t,
lim Ψ (nt) = 0.
n→∞
Hint: Pick Re (σ (A)) < −λ < 0 and use Problem 18 about the spectrum of Ψ (t)
and
for the spectral radius along with Problem 26 to argue that
theorem
Gelfand’s
Ψ (nt) /e−λnt < 1 for all n large enough.
n
P∞
28. Let H be a Hermitian matrix. (H = H ∗ ) . Show that eiH ≡ n=0 (iH)
is unitary.
n!
You have H = U ∗ DU where U is unitary and D is a real diagonal matrix. Then you
have
iλ
e 1
∞
n
X
(iD)
..
eiH = U ∗
U = U∗
U
.
n!
n=0
eiλn
and this is clearly unitary because each matrix in the product is.
29. Show the converse of the above exercise. If V is unitary, then V = eiH for some H
Hermitian.
To do this, note first that V is normal because V ∗ V = V V ∗ = I. Therefore, there
exists a unitary U such that V = U ∗ DU where D is the diagonal matrix consisting
of the eigenvalues of V down the main diagonal. Since V is unitary, it preserves all
lengths and so each diagonal entry of D is of magnitude 1. Thus you have
iλ
e 1
∞
k
X
(iD)
∗
∗
.
.
U
V =U
U = U
.
k!
k=0
eiλn
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
167
Exercises
λ1
where D =
..
.
λn
, each λk real. Then the above equals
∞
k
X
(iU ∗ DU )
k=0
∗
= eiH
k!
where H = U DU is obviously Hermitian.
−1
30. If U is unitary and does not have −1 as an eigenvalue so that (I + U )
that
H = i (I − U ) (I + U )−1
exists, show
is Hermitian. Then, verify that
−1
U = (I + iH) (I − iH)
.
−1
To see this, note first that (I − U ) , (I + U ) commute. Taking the adjoint of the
above operator, it is easy to see that this equals
−i (I + U ∗ )
−1
(I − U ∗ )
−1
= −i (U ∗ U + U ∗ )
−1
= −i (U + I)
−1
= i (I + U )
(U ∗ U − U ∗ )
U U ∗ (U − I)
−1
(I − U ) = i (I − U ) (I + U )
.
Next, using the fact that everything commutes,
−1
−1
−1
I + i i (I − U ) (I + U )
I − i i (I − U ) (I + U )
−1
−1
−1
=
I − (I − U ) (I + U )
I + (I − U ) (I + U )
−1
−1
−1
= (I + U − (I − U )) (I + U )
(I + U + I − U ) (I + U )
−1
−1
−1
−1 1
= 2U (I + U )
2I (I + U )
= 2U (I + U )
(I + U ) = U
2
31. Suppose that A ∈ L (V, V ) where V is a normed linear space. Also suppose that
kAk < 1 where this refers to the operator norm on A. Verify that
(I − A)−1 =
∞
X
Ai
i=0
This is called the Neumann series. Suppose now that you only know the algebraic
−1
condition ρ (A) < 1. Is it still the case that the Neumann series converges to (I − A) ?
Consider partial sums of the series.
q
q
q
X i X
i X
i
≤
A ≤
A
kAk
i=p i=p
i=p
P∞
i
1
< ∞. Therefore, by
which converges to 0 as p, q → ∞ because i=0 kAk = 1−kAk
completeness of L (V, V ) , this series converges. Why is is the desired inverse? This is
obvious when you multiply both sides by (I − A) . Thus
(I − A)
∞
X
i=0
Ai = lim (I − A)
n→∞
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
n
X
i=0
Ai = lim I − An+1 = I
n→∞
7KH6D\ORU)RXQGDWLRQ
168
Exercises
n+1
because An+1 ≤ kAk
and this converges to 0 as n → ∞ because kAk < 1. Yes,
it the series still converges. Let ρ (A) < r < R < 1. Then for all n large enough, you
have
1/n
kAn k
<R
and so kAn k < Rn . It follows that the partial sums of the series form a Cauchy
sequence and so the series converges.
F.34
Exercises
15.3
1. In Example 15.1.10 an eigenvalue was found correct to several decimal places along
with an eigenvector. Find the other eigenvalues along with their eigenvectors.
1 2 3
The matrix was 2 2 1
3 1 4
20
1 2 3
8. 485 7 × 1015 6. 190 4 × 1015 1. 188 8 × 1016
2 2 1 = 6. 190 4 × 1015 4. 515 9 × 1015 8. 672 7 × 1015
3 1 4
1. 188 8 × 1016 8. 672 7 × 1015 1. 665 6 × 1016
0.534 92 −0.470 81 −0.701 57
= 0.390 23 −0.598 82 0.699 38
0.749 39 0.647 89
0.136 60
16
1. 586 4 × 10
1. 157 3 × 1016 2. 222 5 × 1016
0
2. 212 9 × 1011 8. 403 1 × 1011
0
0
4. 160 5 × 1011
T
0.534 92 −0.470 81 −0.701 57
1 2 3
0.390 23 −0.598 82 0.699 38 2 2 1
0.749 39 0.647 89
0.136 60
3 1 4
0.534 92 −0.470 81 −0.701 57
0.390 23 −0.598 82 0.699 38
0.749 39 0.647 89
0.136 60
6. 662 1
1. 218 8 × 10−4 7. 374 9 × 10−5
1. 139 5
−1. 156 9
= 1. 218 8 × 10−4
7. 374 9 × 10−5
−1. 156 9
−0.801 49
0.534 92
The largest eigenvalue is close to 6. 662 1 and an eigenvector is 0.390 23 . The
0.749 39
others can be found by looking at the lower right block. You could do the same as the
above or you could just graph the characteristic function and zoom in on the roots.
They are approximately 1.7 and −1.3. Next use shifted inverse power method to find
them more exactly and to also find the eigenvectors.
−1
1 2 3
1 0 0
−0.977 92 −5. 047 3
3. 47
2 2 1 − 1.7 0 1 0 = −5. 047 3 −33. 47
21. 136
3 1 4
0 0 1
3. 47
21. 136 −13. 281
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
169
Exercises
Rather than use this method as described, I will just use the variation of it based on
the QR algorithm which involves raising the matrix to a power which was just used
to find the first eigenvalue.
20
−0.977 92 −5. 047 3
3. 47
−5. 047 3 −33. 47
21. 136 =
3. 47
21. 136 −13. 281
6. 044 8 × 1031
3. 893 4 × 1032 −2. 458 8 × 1032
3. 893 4 × 1032
2. 507 7 × 1033 −1. 583 7 × 1033 :
32
−2. 458 8 × 10
−1. 583 7 × 1033 1. 000 2 × 1033
0.130 15 −3. 322 8 × 10−2
0.990 94
0.838 32
−0.529 98
−0.127 88
−2
−0.529 42
−0.847 36
4. 112 3 × 10
32
33
4. 644 3 × 10
2. 991 4 × 10
−1. 889 2 × 1033
0
9. 738 8 × 1027 −3. 859 8 × 1028
0
0
3. 342 9 × 1027
T
0.130 15 −3. 322 8 × 10−2
0.990 94
0.838 32
−0.529 98
−0.127 88
−2
−0.529 42
−0.847 36
4. 112 3 × 10
−0.977 92 −5. 047 3
3. 47
−5. 047 3 −33. 47
21. 136
3. 47
21. 136 −13. 281
0.130 15 −3. 322 8 × 10−2
0.990 94
0.838 32
=
−0.529 98
−0.127 88
−2
−0.529 42
−0.847 36
4. 112 3 × 10
−47. 602
6. 755 1 × 10−5 −2. 994 7 × 10−5
6. 755 1 × 10−5 6. 320 5 × 10−2
−0.232 90
−5
−2. 994 7 × 10
−0.232 90
−0.190 38
1
λ−1.7
= −47. 602, Solution is: 1. 679 0 and a
0.130 15
suitable eigenvector is just the first column of the orthogonal matrix 0.838 32 .
−0.529 42
How
well
does
it
work?
1 2 3
0.130 15
0.218 53
2 2 1 0.838 32 = 1. 407 5
3 1 4
−0.529 42
−0.888 91
0.130 15
0.218 52
0.838 32 (1. 679 0) = 1. 407 5
−0.529 42
−0.888 90
Thus you find the eigenvector by solving
This worked well.
Now find the pair which goes with −1.3.
−1
1 2 3
1 0 0
2 2 1 + 1.3 0 1 0
3 1 4
0 0 1
−16. 948 7. 810 9
8. 119 2
= 7. 810 9 −3. 278 5 −3. 802 7
8. 119 2 −3. 802 7 −3. 689 6
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
170
Exercises
20
−16. 948 7. 810 9
8. 119 2
7. 810 9 −3. 278 5 −3. 802 7 =
8. 119 2 −3. 802 7 −3. 689 6
3. 827 4 × 1027 −1. 745 5 × 1027 −1. 823 0 × 1027
−1. 745 5 × 1027 7. 960 3 × 1026
8. 313 7 × 1026 :
27
26
−1. 823 0 × 10
8. 313 7 × 10
8. 682 7 × 1026
0.834 83 −0.544 13 8. 355 9 × 10−2
−0.380 73 −0.461 03
0.801 56
−0.397 63 −0.700 98
−0.592 05
27
4. 584 7 × 10
−2. 090 8 × 1027 −2. 183 7 × 1027
0
1. 658 9 × 1022
2. 792 9 × 1022
0
0
4. 620 1 × 1021
T
0.834 83 −0.544 13 8. 355 9 × 10−2
−0.380 73 −0.461 03
0.801 56
−0.397 63 −0.700 98
−0.592 05
−16. 948 7. 810 9
8. 119 2
7. 810 9 −3. 278 5 −3. 802 7
8. 119 2 −3. 802 7 −3. 689 6
0.834 83 −0.544 13 8. 355 9 × 10−2
−0.380 73 −0.461 03
0.801 56
−0.397 63 −0.700 98
−0.592 05
−24. 377
9. 597 9 × 10−5
5. 267 × 10−5
0.127 02
−1. 802 1 × 10−2
= 9. 597 9 × 10−5
−5
−2
5. 267 × 10
−1. 802 1 × 10
0.334 17
1
λ+1.3
= −24. 377, Solution is: −1. 341. The approximate eigenvector is the first column.
How
well
does it work?
1 2 3
0.834 83
−1. 119 5
2 2 1 −0.380 73 = 0.510 57
3 1 4
−0.397 63
0.533 24
0.834 83
−1. 119 5
−0.380 73 (−1. 341) = 0.510 56
−0.397 63
0.533 22
It works well. You could do more iterations if desired.
3 2 1
2. Find the eigenvalues and eigenvectors of the matrix A = 2 1 3 numerically.
1 3 2
√
In this case the exact eigenvalues are ± 3, 6. Compare with the exact answers.
13
3 2 1
4. 353 6 × 109 4. 353 6 × 109 4. 353 6 × 109
2 1 3 = 4. 353 6 × 109 4. 353 6 × 109 4. 353 6 × 109 =
1 3 2
4. 353 6 × 109 4. 353 6 × 109 4. 353 6 × 109
0.577 35 −0.816 50 −1. 437 8 × 10−29
0.577 35 0.408 25
−0.707 11
0.577 35 0.408 25
0.707 11
7. 540 7 × 109
7. 540 7 × 109
7. 540 7 × 109
0
1. 533 3 × 10−19 1. 533 3 × 10−19
0
0
2. 736 9 × 10−48
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
Exercises
171
T
0.577 35 −0.816 50 −1. 437 8 × 10−29
3 2 1
0.577 35 0.408 25
2 1 3
−0.707 11
0.577 35 0.408 25
0.707 11
1 3 2
−29
0.577 35 −0.816 50 −1. 437 8 × 10
0.577 35 0.408 25
−0.707 11
0.577 35 0.408 25
0.707 11
6. 000 0
0
0
1. 5
0.866 03
= −1. 262 2 × 10−29
−29
−2. 524 4 × 10
0.866 03
−1. 5
0.577 35
An eigenvalue is 6 and an eigenvector is 0.577 35 . Now lets find the others.
0.577 35
−1
3 2 1
1 0 0
2 1 3 − 1.5 0 1 0
1 3 2
0 0 1
2. 740 7
−0.592 59
−1. 925 9
= −0.592 59 7. 407 4 × 10−2 0.740 74
−1. 925 9
0.740 74
1. 407 4
20
2. 740 7
−0.592 59
−1. 925 9
−0.592 59 7. 407 4 × 10−2 0.740 74 =
−1. 925 9
0.740 74
1. 407 4
12
3. 034 1 × 10
−8. 129 9 × 1011 −2. 221 1 × 1012
−8. 129 9 × 1011 2. 178 4 × 1011
5. 951 5 × 1011 :
12
11
−2. 221 1 × 10
5. 951 5 × 10
1. 626 0 × 1012
0.788 68
0.425 47 0.443 81
−0.211 33 −0.490 28 0.845 56
−0.577 35 0.760 66 0.296 76
3. 847 1 × 1012 −1. 030 8 × 1012 −2. 816 3 × 1012
0
3. 850 7 × 106
3. 843 7 × 107
0
0
1. 927 6 × 107
T
0.788 68
0.425 47 0.443 81
2. 740 7
−0.592 59
−1. 925 9
−0.211 33 −0.490 28 0.845 56 −0.592 59 7. 407 4 × 10−2 0.740 74
−0.577 35 0.760 66 0.296 76
−1. 925 9
0.740 74
1. 407 4
0.788 68
0.425 47 0.443 81
−0.211 33 −0.490 28 0.845 56
−0.577 35 0.760 66 0.296 76
4. 309 4
−9. 614 1 × 10−6 −1. 276 5 × 10−5
= −9. 614 1 × 10−6
−0.223 59
0.195 59
−1. 276 5 × 10−5
0.195 59
0.136 42
0.788 68
1
−0.211 33 . How well
λ−1.5 = 4. 309 4, Solution is: 1. 732 1 and an eigenvector is
−0.577 35
does
it
work?
3 2 1
0.788 68
1. 366
2 1 3 −0.211 33 = −0.366 02
1 3 2
−0.577 35
−1.0
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
172
Exercises
0.788 68
1. 366 1
−0.211 33 1. 732 1 = −0.366 04
−0.577 35
−1.0
Next you can find the other one.
−1
3 2 1
1 0 0
2 1 3 + 1.5 0 1 0
1 3 2
0 0 1
−2
4. 444 4 × 10
0.711 11 −0.622 22
0.711 11
−2. 622 2
2. 044 4
=
−0.622 22
2. 044 4
−1. 288 9
20
4. 444 4 × 10−2 0.711 11 −0.622 22
0.711 11
−2. 622 2
2. 044 4 =
−0.622 22
2. 044 4
−1. 288 9
11
2. 178 4 × 10
−8. 129 9 × 1011 5. 951 5 × 1011
−8. 129 9 × 1011 3. 034 1 × 1012 −2. 221 1 × 1012 :
5. 951 5 × 1011 −2. 221 1 × 1012 1. 626 0 × 1012
0.211 32 −0.490 28 0.845 56
−0.788 68 0.425 47 0.443 81
0.577 35
0.760 66 0.296 76
12
1. 030 8 × 10
−3. 847 1 × 1012 2. 816 3 × 1012
0
1. 437 1 × 107
2. 791 7 × 107
0
0
1. 927 6 × 107
T
0.211 32 −0.490 28 0.845 56
−0.788 68 0.425 47 0.443 81
0.577 35
0.760 66 0.296 76
4. 444 4 × 10−2 0.711 11 −0.622 22
0.211 32 −0.490 28 0.845 56
0.711 11
−2. 622 2
2. 044 4 −0.788 68 0.425 47 0.443 81 :
−0.622 22
2. 044 4
−1. 288 9
0.577 35
0.760 66 0.296 76
−4. 309 4
−2. 468 3 × 10−6 −1. 167 8 × 10−5
−2. 468 3 × 10−6
0.280 95
−6. 479 4 × 10−2
−5
−2
−1. 167 8 × 10
−6. 479 4 × 10
0.161 74
0.211 32
1
−0.788 68 . How
λ+1.5 = −4. 309 4, Solution is: −1. 732 1 and an eigenvector is
0.577 35
well
does
it
work?
3 2 1
1 0 0
0.211 32
2 1 3 + 1. 732 1 0 1 0 −0.788 68
1 3 2
0 0 1
0.577 35
−5
−2. 262 8 × 10
= −6. 262 8 × 10−5 which is pretty close to 0 so this worked well.
7. 935 × 10−6
3 2 1
3. Find the eigenvalues and eigenvectors of the matrix A = 2 5 3 numerically.
1 3 2
√
√
The exact eigenvalues are 2, 4 + 15, 4 − 15. Compare your numerical results with
the exact values. Is it much fun to compute the exact eigenvectors?
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
Exercises
173
20
3 2 1
2 5 3 =
1 3 2
1. 448 7 × 1017 2. 713 5 × 1017 1. 632 8 × 1017
2. 713 5 × 1017 5. 082 3 × 1017 3. 058 1 × 1017 =
1. 632 8 × 1017 3. 058 1 × 1017 1. 840 1 × 1017
0.415 99
0.806 58
0.419 97
0.779 18 −7. 803 9 × 10−2 −0.621 92
0.468 86
−0.585 95
0.660 94
17
17
3. 482 5 × 10
6. 522 6 × 10
3. 924 8 × 1017
0
1. 539 × 1013 1. 324 1 × 1013
0
0
1. 399 9 × 1012
T
0.415 99
0.806 58
0.419 97
3 2 1
0.779 18 −7. 803 9 × 10−2 −0.621 92 2 5 3
1 3 2
0.468 86
−0.585 95
0.660 94
0.415 99
0.806 58
0.419 97
0.779 18 −7. 803 9 × 10−2 −0.621 92
0.468 86
−0.585 95
0.660 94
7. 873 0
8. 768 9 × 10−5 3. 295 6 × 10−5
1. 746 2
0.641 05
= 8. 768 9 × 10−5
−5
3. 295 6 × 10
0.641 05
0.380 82
0.415 99
An eigenvalue is 7. 873 0 and the approximate eigenvector is 0.779 18 . How well
0.468 86
does
it
work?
3 2 1
1 0 0
0.415 99
2 5 3 − 7. 873 0 0 1 0 0.779 18
1 3 2
0 0 1
0.468 86
−4
1. 007 3 × 10
= −2. 414 × 10−5
−8. 478 × 10−5
√
4 + 15 = 7. 873 0 which is what was just obtained to four decimal places.
It works pretty well. Next we find the other eigenvalues and eigenvectors which go
with them.
−1
3 2 1
1 0 0
2 5 3 − 1.75 0 1 0 =
1 3 2
0 0 1
3. 295 6 −1. 006 3
−1. 106 9
−1. 006 3 0.276 73
0.704 4
−2
−1. 106 9 0.704 4 −2. 515 7 × 10
20
3. 295 6 −1. 006 3
−1. 106 9
−1. 006 3 0.276 73
0.704 4
−1. 106 9 0.704 4 −2. 515 7 × 10−2
8. 995 9 × 1011 −2. 998 7 × 1011 −2. 998 6 × 1011
= −2. 998 7 × 1011 9. 995 6 × 1010
9. 995 4 × 1010 =
11
10
−2. 998 6 × 10
9. 995 4 × 10
9. 995 2 × 1010
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
174
Exercises
0.904 53 −0.404 72 0.134 23
−0.301 52 −0.829 69 −0.469 80
−0.301 51 −0.384 47 0.872 51
9. 945 4 × 1011 −3. 315 2 × 1011 −3. 315 1 × 1011
0
2. 995 0 × 106
1. 376 × 106
0
0
5. 368 8 × 105
T
0.904 53 −0.404 72 0.134 23
3. 295 6 −1. 006 3
−1. 106 9
−0.301 52 −0.829 69 −0.469 80 −1. 006 3 0.276 73
0.704 4
−2
−0.301 51 −0.384 47 0.872 51
−1. 106 9 0.704 4 −2. 515 7 × 10
0.904 53 −0.404 72 0.134 23
−0.301 52 −0.829 69 −0.469 80 =
−0.301 51 −0.384 47 0.872 51
4. 000 0
1. 828 8 × 10−6 −8. 371 4 × 10−6
1. 828 8 × 10−6
0.155 70
−7. 669 2 × 10−2
−6
−2
−8. 371 4 × 10
−7. 669 2 × 10
−0.608 53
0.904 53
1
−0.301 52 . How well does it
λ−1.75 = 4, Solution is: 2.0. An eigenvector is then
−0.301 51
work?
3 2 1
1 0 0
0.904 53
−0.000 02
2 5 3 − 2 0 1 0 −0.301 52 = −0.000 03
1 3 2
0 0 1
−0.301 51
−0.000 03
−1
3 2 1
1 0 0
2 5 3 − 0.38 0 1 0 =
1 3 2
0 0 1
0.493 54
7. 815 4 × 10−2 −0.449 38
7. 815 4 × 10−2
−1. 056 5
1. 908 3
−0.449 38
1. 908 3
−2. 639 1
20
−2
0.493 54
7. 815 4 × 10
−0.449 38
7. 815 4 × 10−2
−1. 056 5
1. 908 3 =
−0.449 38
1. 908 3
−2. 639 1
7. 593 8 × 109
−4. 459 9 × 1010
6. 738 × 1010
−4. 459 9 × 1010 2. 619 4 × 1011 −3. 957 3 × 1011 :
6. 738 × 1010
−3. 957 3 × 1011 5. 978 7 × 1011
9. 356 7 × 10−2 0.107 43
−0.989 8
−0.549 53
0.834 58 3. 863 5 × 10−2
0.830 22
0.540 31
0.137 12
10
11
8. 115 9 × 10
−4. 766 6 × 10
7. 201 3 × 1011
0
4. 418 3 × 106
1. 380 8 × 106
0
0
6. 663 6 × 105
T
9. 356 7 × 10−2 0.107 43
−0.989 8
−0.549 53
0.834 58 3. 863 5 × 10−2
0.830 22
0.540 31
0.137 12
−2
0.493 54
7. 815 4 × 10
−0.449 38
7. 815 4 × 10−2
−1. 056 5
1. 908 3
−0.449 38
1. 908 3
−2. 639 1
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
175
Exercises
9. 356 7 × 10−2 0.107 43
−0.989 8
−0.549 53
0.834 58 3. 863 5 × 10−2 =
0.830 22
0.540 31
0.137 12
−3. 952 9
−2. 044 4 × 10−5 1. 021 4 × 10−5
−2. 044 4 × 10−5
0.182 25
0.145 62
1. 021 4 × 10−5
0.145 62
0.568 55
1
λ−.38 = −3. 952 9,Solution is: 0.127 02 and an approximate eigenvector is then
9. 356 7 × 10−2
. How well does it work?
−0.549 53
0.830 22
3 2 1
1 0 0
9. 356 7 × 10−2
2 5 3 − 0.127 02 0 1 0
−0.549 53
1 3 2
0 0 1
0.830 22
−2. 388 × 10−5
= −5. 469 9 × 10−5
−3. 754 4 × 10−5
√
Note that 4 − 15 = 0.127 02 which is what was just obtained.
0 2 1
4. Find the eigenvalues and eigenvectors of the matrix A = 2 5 3 numerically.
1 3 2
I don’t know the exact eigenvalues in this case. Check your answers by multiplying
your numerically computed eigenvectors by the matrix.
50
0 2 1
5. 045 4 × 1042 1. 454 4 × 1043 8. 826 9 × 1042
2 5 3 = 1. 454 4 × 1043 4. 192 3 × 1043 2. 544 4 × 1043
1 3 2
8. 826 9 × 1042 2. 544 4 × 1043 1. 544 2 × 1043
0.284 32 0.756 78
0.588 6
= 0.819 59 −0.510 39 0.260 32 ·
0.497 42 0.408 40 −0.765 36
1. 774 5 × 1043 5. 115 1 × 1043 3. 104 5 × 1043
0
7. 091 8 × 1038 8. 072 9 × 1037
0
0
3. 007 5 × 1038
A matrix similar to the original is
0.284 32 0.756 78
0.588 6
0.819 59 −0.510 39 0.260 32
0.497 42 0.408 40 −0.765 36
0.284 32 0.756 78
0.588 6
0.819 59 −0.510 39 0.260 32
0.497 42 0.408 40 −0.765 36
7. 514 5
9. 130 9 × 10−5
9. 130 9 × 10−5
−0.541 46
1. 248 7 × 10−5
−0.344 28
T
0
2
1
2 1
5 3
3 2
=
1. 248 7 × 10−5
−0.344 28
−2
2. 686 9 × 10
0.284 32
Thus an eigenvalue is 7. 514 5 and an eigenvector is approximately 0.819 59 . How
0.497 42
well does it work?
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
176
Exercises
0 2 1
0.284 32
2. 136 6
2 5 3 0.819 59 = 6. 158 9
1 3 2
0.497 42
3. 737 9
0.284 32
2. 136 5
0.819 59 7. 514 5 = 6. 158 8 .
0.497 42
3. 737 9
This has essentially found it. However, we don’t know the others yet. begin with the
one which is closest to −0.541 46
−1
0 2 1
1 0 0
2 5 3 + .541 46 0 1 0
1 3 2
0 0 1
−5. 323 8
2. 181 4
−0.480 23
−0.393 89 −0.393 38
= 2. 181 4
−0.480 23 −0.393 38
1. 046 8
20
−5. 323 8
2. 181 4
−0.480 23
2. 181 4
−0.393 89 −0.393 38 =
−0.480 23 −0.393 38
1. 046 8
15
5. 483 4 × 10
−2. 055 8 × 1015 2. 530 4 × 1014
−2. 055 8 × 1015 7. 707 7 × 1014 −9. 487 0 × 1013 :
2. 530 4 × 1014 −9. 487 0 × 1013 1. 167 7 × 1013
0.935 48
0.353 06
1. 491 6 × 10−2
−0.350 73
0.932 81
−8. 286 3 × 10−2
−2
−2
4. 316 9 × 10
−7. 228 5 × 10
−0.996 45
15
15
5. 861 6 × 10
−2. 197 6 × 10
2. 704 9 × 1014
0
2. 167 1 × 1010 −1. 796 2 × 109
0
0
8. 270 5 × 107
T
0.935 48
0.353 06
1. 491 6 × 10−2
−0.350 73
0.932 81
−8. 286 3 × 10−2
−2
−2
4. 316 9 × 10
−7. 228 5 × 10
−0.996 45
−5. 323 8
2. 181 4
−0.480 23
2. 181 4
−0.393 89 −0.393 38
−0.480 23 −0.393 38
1. 046 8
0.935 48
0.353 06
1. 491 6 × 10−2
−0.350 73
0.932 81
−8. 286 3 × 10−2 =
−2
−2
4. 316 9 × 10
−7. 228 5 × 10
−0.996 45
−5
−6. 163 8
1. 742 0 × 10
−1. 159 3 × 10−6
1. 742 0 × 10−5
0.513 51
0.577 10
−1. 159 3 × 10−6
0.577 10
0.979 41
1
λ+.541 46
= −6. 163 8, Solution is: −0.703 70
0.935 48
. How well does it work?
−0.350 73
−2
4. 316 9 × 10
0 2 1
1 0 0
0.935 48
2 5 3 + 0.703 70 0 1 0
:
−0.350 73
−2
1 3 2
0 0 1
4. 316 9 × 10
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
177
Exercises
6. 276 × 10−6
8. 299 × 10−6
6. 025 3 × 10−6
Next find the eigenvalue closest to 0.
−1
0 2 1
−1.0 1.0 −1.0
2 5 3 = 1.0
1.0 −2.0
1 3 2
−1.0 −2.0 4.0
20
−1.0 1.0 −1.0
1.0
1.0 −2.0
−1.0 −2.0 4.0
1. 287 1 × 1013
2. 778 8 × 1013 −5. 314 3 × 1013
5. 999 6 × 1013 −1. 147 4 × 1014 :
= 2. 778 8 × 1013
13
−5. 314 3 × 10
−1. 147 4 × 1014 2. 194 2 × 1014
0.209 85 −0.930 85 0.299 13
0.453 05 −0.178 55 −0.873 42
−0.866 43 −0.318 81 −0.384 26
6. 133 5 × 1013 1. 324 3 × 1014 −2. 532 5 × 1014
0
1. 509 3 × 109
1. 706 9 × 109
0
0
6. 192 3 × 109
T
0.209 85 −0.930 85 0.299 13
0.453 05 −0.178 55 −0.873 42
−0.866 43 −0.318 81 −0.384 26
−1.0 1.0 −1.0
0.209 85 −0.930 85 0.299 13
1.0
1.0 −2.0 0.453 05 −0.178 55 −0.873 42 =
−1.0 −2.0 4.0
−0.866 43 −0.318 81 −0.384 26
5. 288 0
2. 017 9 × 10−5 −2. 856 3 × 10−5
2. 017 9 × 10−5
−0.916 86
0.727 58
−5
−2. 856 3 × 10
0.727 58
−0.371 12
0.209 85
1
0.453 05 .
λ = 5. 288 0, Solution is: 0.189 11. An approximate eigenvector is
−0.866 43
How well does it work?
0 2 1
1 0 0
0.209 85
−1. 473 4 × 10−5
2 5 3 − 0.189 11 0 1 0 0.453 05 = −1. 628 6 × 10−5
1 3 2
0 0 1
−0.866 43
−9. 422 7 × 10−6
.
0 2 1
5. Find the eigenvalues and eigenvectors of the matrix A = 2 0 3 numerically.
1 3 2
I don’t know the exact eigenvalues in this case. Check your answers by multiplying
your numerically computed eigenvectors by the matrix.
I will do this one a little differently. You can easily find that the eigenvalues are
approximately 4.9, −2.7, .3. To find the eigen pair which goes with .3 to the following.
−1
0 2 1
1 0 0
2 0 3 − .3 0 1 0 =
1 3 2
0 0 1
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
178
Exercises
−1. 138 5
−4. 788 7 × 10−2 0.754 22
−4. 788 7 × 10−2
−0.180 77
0.347 18
0.754 22
0.347 18
−0.468 10
20
−1. 138 5
−4. 788 7 × 10−2 0.754 22
−4. 788 7 × 10−2
−0.180 77
0.347 18 =
0.754 22
0.347 18
−0.468 10
17805.
3431. 1 −12214.
3431. 1
661. 21 −2353. 7 =
−12214. −2353. 7 8378. 3
0.814 41 −0.321 53 −0.483 08
0.156 94
0.923 48 −0.350 07
−0.558 67 −0.209 28 −0.802 55
21863.
4213.0
−14997.
0
2. 261 7 × 10−2 6. 348 1 × 10−2
0
0
0.281 26
A similar matrix is
T
0.814 41 −0.321 53 −0.483 08
0.156 94
0.923 48 −0.350 07
−0.558 67 −0.209 28 −0.802 55
−1. 138 5
−4. 788 7 × 10−2 0.754 22
−4. 788 7 × 10−2
−0.180 77
0.347 18
0.754 22
0.347 18
−0.468 10
0.814 41 −0.321 53 −0.483 08
0.156 94
0.923 48 −0.350 07 =
−0.558 67 −0.209 28 −0.802 55
−1. 665 1
7. 437 4 × 10−6 1. 163 3 × 10−5
7. 437 4 × 10−6
−0.296 62
−0.142 05
−5
1. 163 3 × 10
−0.142 05
0.174 36
Thus an eigenvalue is close to −1.6651 and so the eigenvalue for the original matrix
1
is the solution of λ−.3
= −1.6651, Solution is: −0.300 56. Then what about the
eigenvector? We have
1
(A − .3I)−1 v =
v
λ − .3
It follows that (λ − .3) v = (A − .3I) v. Thus you have Av = λv. Thus an eigenvector
is just the first column of that orthogonal matrix above. Check it.
0 2 1
0.814 41
−0.244 79
2 0 3 0.156 94 = −0.047 19
1 3 2
−0.558 67
0.167 89
0.814 41
−0.244 78
0.156 94 (−0.300 56) = −4. 717 0 × 10−2
−0.558 67
0.167 91
This worked very well. Now lets find the pair associated with the eigenvalue being
close to −2.7.
−1
0 2 1
1 0 0
2 0 3 + 2.7 0 1 0
1 3 2
0 0 1
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
Exercises
7. 969 8 −13. 823 7. 127 4
= −13. 823 25. 248 −13. 175
7. 127 4 −13. 175 7. 105 8
20
7. 969 8 −13. 823 7. 127 4
−13. 823 25. 248 −13. 175 =
7. 127 4 −13. 175 7. 105 8
1. 896 8 × 1031 −3. 436 7 × 1031 1. 799 7 × 1031
−3. 436 7 × 1031 6. 226 6 × 1031 −3. 260 7 × 1031 =
1. 799 7 × 1031 −3. 260 7 × 1031 1. 707 5 × 1031
0.439 25
−0.891 55
0.110 45
−0.795 85
−0.443 21
−0.412 55
0.416 76 9. 330 8 × 10−2 −0.904 21
4. 318 3 × 1031 −7. 823 9 × 1031 4. 097 2 × 1031
0
7. 478 5 × 1026 −3. 772 7 × 1026
0
0
3. 494 4 × 1026
T
0.439 25
−0.891 55
0.110 45
−0.795 85
−0.443 21
−0.412 55
0.416 76 9. 330 8 × 10−2 −0.904 21
7. 969 8 −13. 823 7. 127 4
0.439 25
−0.891 55
−13. 823 25. 248 −13. 175 −0.795 85
−0.443 21
7. 127 4 −13. 175 7. 105 8
0.416 76 9. 330 8 × 10−2
39. 777
−1. 791 2 × 10−4 6. 881 × 10−5
−1. 791 2 × 10−4
0.336 06
−0.128 95
−5
6. 881 × 10
−0.128 95
0.210 75
179
0.110 45
−0.412 55 =
−0.904 21
1
Eigenvalue, λ+2.7
= 39. 777, Solution is: −2. 674 9. The eigenvector would be the first
column of the above orthogonal matrix on the right.
0 2 1
0.439 25
−1. 174 9
2 0 3 −0.795 85 = 2. 128 8
1 3 2
0.416 76
−1. 114 8
0.439 25
−1. 174 9
−0.795 85 (−2. 674 9) = 2. 128 8 This worked very well.
0.416 76
−1. 114 8
Now consider the largest eigenvalue.
−1
0 2 1
1 0 0
2 0 3 − 4.9 0 1 0
1 3 2
0 0 1
1. 753 6 2. 962 0 3. 668 8
= 2. 962 0 4. 446 3 5. 621
3. 668 8 5. 621 6. 735 1
40
1. 753 6 2. 962 0 3. 668 8
2. 962 0 4. 446 3 5. 621 =
3. 668 8 5. 621 6. 735 1
1. 144 5 × 1044 1. 765 1 × 1044 2. 164 3 × 1044
1. 765 1 × 1044 2. 722 1 × 1044 3. 337 8 × 1044
2. 164 3 × 1044 3. 337 8 × 1044 4. 092 8 × 1044
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
180
Exercises
0.379 20
0.739 6
−0.556 06
0.584 81 −0.657 26 −0.475 40
0.717 08 0.144 92
0.681 76
44
3. 018 2 × 10
4. 654 7 × 1044
0
6. 539 3 × 1039
0
0
0.379 20
0.739 6
−0.556 06
0.584 81 −0.657 26 −0.475 40
0.717 08 0.144 92
0.681 76
1. 753 6 2. 962 0 3. 668 8
2. 962 0 4. 446 3 5. 621
3. 668 8 5. 621 6. 735 1
0.379 20
0.739 6
−0.556 06
0.584 81 −0.657 26 −0.475 40
0.717 08 0.144 92
0.681 76
13. 259
1. 247 5 × 10−4
1. 247 5 × 10−4
−0.142 61
1. 129 2 × 10−5 2. 290 2 × 10−2
1
λ−4.9
5. 707 5 × 1044
5. 579 8 × 1039
4. 797 6 × 1039
T
=
1. 129 2 × 10−5
2. 290 2 × 10−2
−0.181 74
= 13. 259, Solution is: 4. 975 4. Check this.
0 2 1
0.379 20
1. 886 7
2 0 3 0.584 81 = 2. 909 6
1 3 2
0.717 08
3. 567 8
0.379 20
1. 886 7
0.584 81 4. 975 4 = 2. 909 7 . It worked well.
0.717 08
3. 567 8
3 2 3
T
6. Consider the matrix A = 2 1 4 and the vector (1, 1, 1) . Find the shortest
3 4 0
distance between the Rayleigh quotient determined by this vector and some eigenvalue
of A.
T
1
3 2 3
1
q = 1 2 1 4 1 31 = 7. 333 3
1
3 4 0
1
3 2 3
1
1 2 1 4 1 − 7. 333 3 1
3 4 0
1
1 √
|7. 333 3 − λq | ≤
3
= 0.471 41
Thus there is an eigenvalue close to 7.333.
1 2 1
T
7. Consider the matrix A = 2 1 4 and the vector (1, 1, 1) . Find the shortest
1 4 5
distance between the Rayleigh quotient determined by this vector and some eigenvalue
of A.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
181
Exercises
T
1
1
q= 1 2
1
1
2 1
1
1 4 1 13 = 7
4 5
1
|7 − λq | = 2. 449 5
3 2 3
T
8. Consider the matrix A = 2 6 4 and the vector (1, 1, 1) . Find the shortest
3 4 −3
distance between the Rayleigh quotient determined by this vector and some eigenvalue
of A.
T
1
3 2 3
1
q = 1 2 6 4 1 13 = 8.0
1
3 4 −3
1
|λq − 8| ≤
=
9.
Using
3
2
3
3
2
3
3. 266 0
2
6
4
3
1
1 4 1 − 8 1
−3
1
1 √
3
Gerschgorin’s
theorem, find upper and lower bounds for the eigenvalues of A =
2 3
6 4 .
4 −3
−10 ≤ λ ≤ 12
10. Tell how to find a matrix whose characteristic polynomial is a given monic polynomial.
This is called a companion matrix. Find the roots of the polynomial x3 + 7x2 + 3x + 7.
0 0 −7
The companion matrix is 1 0 −3
0 1 −7
20
0 0 −7
8. 062 3 × 1014 −5. 408 5 × 1015 3. 628 2 × 1016
1 0 −3 = 2. 253 5 × 1014 −1. 511 7 × 1015 1. 014 1 × 1016
0 1 −7
7. 726 4 × 1014 −5. 183 2 × 1015 3. 477 × 1016
0.707 72 0.258 58
0.657 47
= 0.197 82 0.820 86 −0.535 78 ·
0.678 23 −0.509 24 −0.529 79
1. 139 2 × 1015 −7. 642 2 × 1015 5. 126 6 × 1016
0
4. 570 4 × 1010 2. 079 1 × 1010
0
0
3. 151 4 × 1011
T
0.707 72 0.258 58
0.657 47
0 0 −7
0.197 82 0.820 86 −0.535 78 1 0 −3 ·
0.678 23 −0.509 24 −0.529 79
0 1 −7
0.707 72 0.258 58
0.657 47
0.197 82 0.820 86 −0.535 78
0.678 23 −0.509 24 −0.529 79
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
182
Exercises
−6. 708 3
= 4. 147 2 × 10−5
−1. 391 6 × 10−5
5. 850 6
5. 220 9
0.154 76
1. 187 6
−0.936 81 −0.446 46
Clearly a real root is close to −6. 708 3. Then the other roots can be obtained from
the lower right block.
0.154 76
1. 187 6
, eigenvalues: −0.145 85 + 1. 011i, −0.145 85 − 1. 011i
−0.936 81 −0.446 46
How well do these work? Try the last one. Evaluating the polynomial at this value of
x gives
6. 350 5 × 10−4 + 2. 065 8 × 10−4 i
the roots to x4 + 3x3 + 4x2 + x + 1. It has two complex roots.
20
0 0 −1
0 0 −1
=
1 0 −4
0 1 −3
36482.
−27300.0
−49899.
2. 448 9 × 105
14010.0
9182.0
−77199.
1. 949 9 × 105
1. 318 0 × 105 −95190.0 −1. 904 1 × 105 9. 023 5 × 105 =
27300.0
49899.
−2. 448 9 × 105 5. 442 5 × 105
0.260 30
−6. 965 1 × 10−2
−0.941 85
0.200 79
9. 996 0 × 10−2
0.251 72
0.209 32
0.939 59
0.940 38
−0.202 95
0.253 14
−0.102 07
0.194 78
0.943 71
−7. 090 6 × 10−2 −0.257 75
1. 401 6 × 105 −85984. −2. 474 6 × 105 1. 037 8 × 106
0
70622. −2. 084 2 × 105 3. 625 1 × 105
0
0
2. 130 5
−5. 038
0
0
0
2. 683 4
T
0.260 30
−6. 965 1 × 10−2
−0.941 85
0.200 79
9. 996 0 × 10−2
0.251 72
0.209 32
0.939 59
0.940 38
−0.202 95
0.253 14
−0.102 07
0.194 78
0.943 71
−7. 090 6 × 10−2 −0.257 75
0 0 0 −1
1 0 0 −1
0 1 0 −4
0 0 1 −3
0.260 30
−6. 965 1 × 10−2
−0.941 85
0.200 79
9. 996 0 × 10−2
0.251 72
0.209 32
0.939 59
=
0.940 38
−0.202 95
0.253 14
−0.102 07
0.194 78
0.943 71
−7. 090 6 × 10−2 −0.257 75
−0.613 47
−4. 251 0
0.485 69
2. 096 8
0.503 89
−2. 337 6
0.115 42 0.330 94
2. 547 6 × 10−7 8. 418 7 × 10−6 −0.157 34 0.304 46
4. 611 8 × 10−6 7. 585 5 × 10−6 −0.974 48 0.108 46
11. Find
0
1
0
0
Now there are two blocks of importance.
−0.613 47 −4. 251 0
, eigenvalues: −1. 475 5 + 1. 182 7i, −1. 475 5 − 1. 182 7i
0.503 89 −2. 337 6
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
183
Exercises
−0.157 34 0.304 46
−0.974 48 0.108 46
, eigenvalues: −0.024 44 + 0.528 23i, −0.024 44 − 0.528 23i
How well does one of these work? Try one.
4
3
(−0.024 44 + 0.528 23i) + 3 (−0.024 44 + 0.528 23i)
2
+4 (−0.024 44 + 0.528 23i) + (−0.024 44 + 0.528 23i)
+1 = 2. 887 9 × 10−5 − 3. 009 2 × 10−6 i
12. Suppose A is a real symmetric matrix and the technique of reducing to an upper
Hessenberg matrix is followed. Show the resulting upper Hessenberg matrix is actually
equal to 0 on the top as well as the bottom.
Let QT AQ = H where H is upper Hessenberg. Then take the transpose of both sides.
This will show that H = H T and so H is zero on the top as well.
6D\ORU85/KWWSZZZVD\ORURUJFRXUVHVPD
7KH6D\ORU)RXQGDWLRQ
© Copyright 2024