11.2MH1 Linear Algebra
1
2
1
1
1.
1
2
1
1
2
2
2
4
1
2
1
1
1
0
0
0
1
0
0
0
2
1
4
2
1
0
0
0
2
2
8
4
1
0
0
0
1
1
0
0
2
0
1
0
1
0
0
1
2
1
0
0
0
1
1
0
0
0
0
1
0
0 2
0
0
1
2
2
2
0
6
6
Sample Examination
r2
r3
r4
r3
r4
r1
r1
So the general solution is: x
r3
r4
r1
r1
2
r2
r3
r4
1
0
0
0
2r1
r1
r1
1
0
0
0
4r2
2r2
1
0
0
0
r4
1
0
0
0
1
1
0
0
r2
1
0
0
0
0
1
0
0
,y
1
0
0
0
2
1
0
0
2
0
1
0
0
0
1
0
,z
2
2
4
2
1
0
0
0
0
0
0
1
1
0
0
0
2
4
8
4
2
2
0
0
1
2 r2
r2
remove rows of 0s
add parameter rows
2
2
0 2
0
0
1
2, w
Solutions
r1
r1
2r3
2
.
[10 marks]
2.
(a)
i. Let 0W be the zero vector of W . Then 0W U and 0W V since U and V
subspaces. Hence 0W U V .
ii. If w1 w2 U V , then w1 w2 U and w1 w2 V . Thus w1 w2 U since U
subspace of W , and similarly w1 w2 V . Therefore w1 w2 U V .
iii. If w U V and
, then w U since U is a subspace, and similarly w
Hence w U V .
By the subspace test, U V is a subspace of W .
2, U
(b) There are many possible examples. Here is a simple one: Let W
2 y 0 , and V
2 x 0 . Then 1 0
xy
0 1 U V , but 1 0
1 1 U V . Hence U V violates the subspace test, so is not a subspace.
are
is a
V.
[6]
xy
01
[4]
3.
i. v1
vn are linearly independent if, whenever
1 v1
1
n vn
are such that
n
0
then
1
0
n
[1]
ii. v1
vn is a spanning set for V if for every element v V there exist
such that
v
n vn
1 v1
1
n
[1]
(a) Assumse that
with
v1
v3
v2
v3
v1
2v3
0
Then
v1
v2
2 v3
0
and since v1 v2 v3 are linearly independent it follows that
2
0
But the only solution to this system of equations in
is the trivial one
Hence v1 v3 v2 v3 v1 2v3 are linearly independent.
(b) Since v1 v2 v3 is a spanning set for V , there are scalars
v4
v1
v2
0.
[6]
such that
v3
Hence
v1
v2
and so v1 v2 v3 v4 are linearly dependent.
v3
1 v4
0
[4]
4.
(a) U is the row space of a 4
row echelon form:
1
3
1
2
2
5
0
2
1
4
3
4
1
2
1
0
4 matrix, and we can find a basis by putting this matrix into
r2
r3
r4
r2
r3
r4
1
0
0
0
3r1
r1
2r1
1
0
0
0
so
1211
0
11
1
2
1
2
2
2
1
0
0
1
1
0
0
1
1
2
2
1
1
2
2
r2
r4
r3
r4
r3 2r2
2r2
1
1
0
0
is a basis for the row space.
[8]
(b) From the above calculation, we can see that
u3
u1
02
22
u4
2u1
Hence
u1
u3
u4
0000
[3]
5.
(a) Find the general solution of Ax
1
2
1
3
2
5
1
5
1
1
1
1
3
5
5
11
1
0
0
0
2
1
0
0
1
1
3
3
3
1
1
1
1
0
0
0
2
1
0
0
1
1
1
0
3 0
1 0
1
3 0
0 0
1
0
0
0
2
1
0
0
1
1
1
0
0
0
0
1
0
0
0
0
0
0
0
0
0:
r2
r3
r4
r3
r2
r3
r4
1
0
0
0
1
3 r3
2
1
0
0
r1
r2
r1
r2
1
0
0
0
r3
r3
1
0
0
0
0
1
0
0
0
0
1
0
2
1
1
1
1
1
1
3
remove row of 0 s
insert parameter row
3
3
1
0
0
0
2r1
r1
3r1
2
1
0
0
0
0
0
1
1
1
2
2
3
1
2
2
3 0
1 0
1
3 0
1 0
0
0
0
0
r3
r4
r4
1
0
0
0
2
1
0
0
1
1
1
0
0
0
1
0
0
0
0
1
10
3
4
3
3
1
1
3
r4
0
0
0
1
r1
r3
r4
r2
r2
3r3
r1
r2
r3
r1
r1
r2
r3
3r4
r4
1
3 r4
2r2
3
6
4
3
3
Hence the general solution is
6
x
4
3
3
and
18
4
1
3
is a basis for the solution space.
(b) The solution space has dimension 1. The rank of A is 3.
[8]
[2]
(c) The first three columns of A are linearly independent, so form a basis for the column
space of A.
[4]
6.
(a) Let
m
v
u
u
i
i
i 1
Then for each j
1
m we have
m
v
u
j
u
j
i
i
u
j
u
j
j
0
i 1
Hence v is orthogonal to each
(b) Put u1
12
j,
1 1 . Then u1
2 3 1 0 . Then u2 v1
u2
Then u2
1 2
7 7
1
7
1
7
7v1
u2
u1
112
1
7, so
31
3
1
u1
7
1
7
1 1
7 7
1 . Then u3 v1
u3
Then u3
152
u3
62
v3
7v1
122
152
2
7
1
7
1
,
7
7 and u3 v2
1
v2
7
15 6
7 7
3 10
,
7
7
u3
3 10
12
7
so put
15
7
so
5
10
2
10
4
10
5
10
Then v1 v2 v3 is an orthonormal basis for U.
(c) Put u
[5]
7, so put
u2
v2
Put u3
vm .
7, so put
1
u1
7
v1
Put u2
and hence to Span v1
[8]
1 0 0 0 and
3
v
u
u vi vi
i 1
5 25 3
10 1
7 10 7
10 7
Then v is orthogonal to U, by part (a).
u
1
v1
7
20
0
10
1
v2
7
25
10
5
v3
10
25 4 13 5
14 7 7 2
[4]
7.
(a)
(b)
i. The null space N T of T is the subset v V T v
0 of V .
ii. The range space R T of T is the subset T v v V of W .
[1]
[1]
i. The 0-vector 0W of W belongs to R T , since 0W T 0V , where 0V is the 0-vector
in V .
ii. Let w1 w2 R T . Then there are elements v1 v2 V such that T v1
w1 and
T v2
w2 . Since T is a linear transformation, we have T v1 v2
T v1
T v2
w1 w2 . Hence w1 w2 R T .
iii. Let w T v R T and
. Then w
T v
T v RT .
By the subspace test, R T is a subspace of W .
[5]
8.
(a) rank T
dim V
nullity T
5
2
3, by the Rank-Nullity Theorem.
(b) No. The range space of T has dimension rank T
whole of W .
3
4
[2]
dim W , so R T is not the
[4]
9.
(a) The eigenvalues of A are the roots of the characteristic polynomial
det I
A
Hence the eigenvalues are
(b) The
1
1
1
0
1
0
0
2
1
1
1 (with multiplicity 2) and
1-eigenspace is the solution space of the system I
0 0
1 0
1 0
0 0
2 0
2 0
1 0 2
0 1 0
0 0 1
0
2
1
1 (with multiplicity 1).
AX
[3]
0.
1 0 0
0 1 0
0 0 1
2
Hence the solution space has dimension 2 and basis
2
0
1
0
1
0
[4]
The
1-eigenspace is the solution space of the system I
2 0 0 0
1 2 2 0
1 0 0 0
1 0 0
0 1 1
0 0 1
0
0
AX
1 0 0
0 1 0
0 0 1
0.
0
Hence the solution space has dimension 1 and basis
0
1
1
[4]
The three eigenvectors
2
0
1
0
1
0
are linearly independent, and so form a basis for
0
1
1
3.
Hence A is diagonalisable.
[2]