Eigenvalues and Eigenvectors
MATH 322, Linear Algebra I
J. Robert Buchanan
Department of Mathematics
Spring 2015
Objectives
In this lesson we will define the terms eigenvalue and
eigenvector and discuss some of their basic properties.
Eigensystems
Definition
If A is an n × n matrix and if there exists a nonzero vector
x ∈ Rn and a scalar λ such that Ax = λ x then λ is an
eigenvalue of A (or of TA ) and x is said to be an eigenvector
corresponding to λ.
Eigensystems
Definition
If A is an n × n matrix and if there exists a nonzero vector
x ∈ Rn and a scalar λ such that Ax = λ x then λ is an
eigenvalue of A (or of TA ) and x is said to be an eigenvector
corresponding to λ.
Remarks:
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The zero vector 0 is never an eigenvector since A 0 = λ 0
for all λ.
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Geometrically an eigenvalue/eigenvector pair means A x is
a dilation/reflection of x by scalar λ.
Example
3
0
1
Suppose A =
, then x =
is an eigenvector
8 −1
2
corresponding to λ = 3 since
3
0
1
3
Ax =
=
= 3x.
8 −1
2
6
Finding Eigenvalues
We must find nonzero solutions of
Ax = λx
Ax − λx = 0
λx − Ax = 0
λI x − A x = 0
(λI − A)x = 0
=⇒ det(λI − A) = 0
Remarks:
I
The expression det(λI − A) is called the characteristic
polynomial of A.
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The expression det(λI − A) = 0 is called the
characteristic equation of A.
Result
Theorem
If A is an n × n matrix, then λ is an eigenvalue of A if and only if
it satisfies the characteristic equation of A
det(λI − A) = 0.
Example
Find all the eigenvalues of A =
3
0
.
8 −1
Example
Find all the eigenvalues of A =
3
0
.
8 −1
det(λI − A) = 0
λ−3
0 −8 λ + 1 = 0
(λ − 3)(λ + 1) = 0
λ = 3 or λ = −1
Characteristic Polynomial and Equation
If A is an n × n matrix then
det(λI − A) = λn + c1 λn−1 + · · · + cn = p(λ)
where p is a polynomial.
Characteristic Polynomial and Equation
If A is an n × n matrix then
det(λI − A) = λn + c1 λn−1 + · · · + cn = p(λ)
where p is a polynomial.
Observation: An n × n matrix will have n eigenvalues. Some of
these may be complex numbers.
Example


2 −3 1
Find the eigenvalues of A =  1 −2 1 .
1 −3 2
Solution
det(λI − A) = 0
λ−2
3
−1
−1 λ + 2
−1 = 0
−1
3 λ−2 Solution
det(λI − A) = 0
λ−2
3
−1
−1 λ + 2
−1 = 0
−1
3 λ−2 λ3 − 2λ2 + λ = 0
λ(λ − 1)2
= 0
λ = 0 or λ = 1
Eigenvalues and Triangular Matrices
Theorem
If A is an n × n triangular matrix (upper triangular, lower
triangular, or diagonal), then the eigenvalues of A are the
diagonal entries of A.
Eigenvalues and Triangular Matrices
Theorem
If A is an n × n triangular matrix (upper triangular, lower
triangular, or diagonal), then the eigenvalues of A are the
diagonal entries of A.
Proof.
The determinant of a triangular matrix is the product of the
diagonal entries.
Example
Find the eigenvalues of

3
 0

 0
0
1
8
0
0

7 11
1 12 
.
0 2 
0 4
Example
Find the eigenvalues of

3
 0

 0
0
1
8
0
0

7 11
1 12 
.
0 2 
0 4
λ1 = 3
λ2 = 8
λ3 = 0
λ4 = 4
Summary
Theorem
If A is an n × n matrix and λ ∈ R, then the following are
equivalent.
1. λ is an eigenvalue of A.
2. The system of equations (λI − A)x = 0 has nontrivial
solutions.
3. There exists a nonzero vector x ∈ Rn such that Ax = λx.
4. λ is a solution of the characteristic equation
det(λI − A) = 0.
Eigenspaces
Remark: The eigenvectors of A x = λ x form a vector space
called the eigenspace.
Eigenspaces
Remark: The eigenvectors of A x = λ x form a vector space
called the eigenspace.
Remarks: the eigenspace is
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the null space of λ I − A,
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the kernel of the matrix transformation TλI−A : Rn → Rn ,
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the set of vectors for which A x = λ x.
Example


2 −3 1
Previously we found the eigenvalues of A =  1 −2 1  to
1 −3 2
be λ1 = 0 and λ2 = λ3 = 1. Find the bases of the eigenspaces
corresponding to each eigenvalue.
Solution (1 of 2)
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Take λ1 = 0. In this case


−2 3 −1
λ1 I − A = −A =  −1 2 −1  .
−1 3 −2
Solution (1 of 2)
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Take λ1 = 0. In this case


−2 3 −1
λ1 I − A = −A =  −1 2 −1  .
−1 3 −2
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If we row reduce this matrix we have


1 0 −1
λ1 I − A ∼  0 1 −1  .
0 0
0
Solution (1 of 2)
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Take λ1 = 0. In this case


−2 3 −1
λ1 I − A = −A =  −1 2 −1  .
−1 3 −2
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If we row reduce this matrix we have


1 0 −1
λ1 I − A ∼  0 1 −1  .
0 0
0
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A basis for the eigenspace is {(1, 1, 1)}.
Solution (1 of 2)
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Take λ2 = λ3 = 1. In this case


−1 3 −1
λ2 I − A = I − A  −1 3 −1  .
−1 3 −1
Solution (1 of 2)
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Take λ2 = λ3 = 1. In this case


−1 3 −1
λ2 I − A = I − A  −1 3 −1  .
−1 3 −1
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If we row reduce this matrix we have


1 −3 1
0 0 .
λ2 I − A ∼  0
0
0 0
Solution (1 of 2)
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Take λ2 = λ3 = 1. In this case


−1 3 −1
λ2 I − A = I − A  −1 3 −1  .
−1 3 −1
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If we row reduce this matrix we have


1 −3 1
0 0 .
λ2 I − A ∼  0
0
0 0
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A basis for the eigenspace is {(1, 3, 0), (1, 0, −1)}.
Powers of a Matrix
Theorem
If k is a positive integer and λ is an eigenvalue of A with
corresponding eigenvector x, then λk is an eigenvalue of Ak
with x as its corresponding eigenvector.
Powers of a Matrix
Theorem
If k is a positive integer and λ is an eigenvalue of A with
corresponding eigenvector x, then λk is an eigenvalue of Ak
with x as its corresponding eigenvector.
Proof.
Suppose A x = λ x then
Ak x = Ak −1 (A x)
= Ak −1 λ x
= λ Ak −2 (A x)
= λ2 Ak −2 x
..
.
Ak x = λk x.
Eigenvalues and Invertibility
Theorem
A square matrix A is invertible if and only if 0 is not an
eigenvalue of A.
Proof
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Note that if A is an n × n matrix then λ = 0 is a solution to
the characteristic equation
λn + c1 λn−1 + · · · + cn−1 λ + cn = 0
if and only if cn = 0.
Proof
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Note that if A is an n × n matrix then λ = 0 is a solution to
the characteristic equation
λn + c1 λn−1 + · · · + cn−1 λ + cn = 0
if and only if cn = 0.
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Recall that
det(λ I − A) = λn + c1 λn−1 + · · · + cn−1 λ + cn
so that
det((0) I − A) = det(−A) = (−1)n det(A) = cn .
Proof
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Note that if A is an n × n matrix then λ = 0 is a solution to
the characteristic equation
λn + c1 λn−1 + · · · + cn−1 λ + cn = 0
if and only if cn = 0.
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Recall that
det(λ I − A) = λn + c1 λn−1 + · · · + cn−1 λ + cn
so that
det((0) I − A) = det(−A) = (−1)n det(A) = cn .
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Thus det(A) = 0 if and only if cn = 0. Equivalently A is
invertible if and only if cn 6= 0.
Equivalent Statements
Theorem
If A is an n × n matrix and if TA : Rn → Rn is multiplication by A,
then the following are equivalent.
1. A is invertible.
2. A x = 0 has only the trivial solution.
3. The reduced row echelon form of A is In .
4. A is expressible as the product of elementary matrices.
5. A x = b is consistent for every n × 1 matrix b.
6. A x = b has exactly one solution for every n × 1 matrix b.
7. det(A) 6= 0.
8. The column vectors of A are linearly independent.
9. The row vectors of A are linearly independent.
10. The column vectors of A span Rn .
Equivalent Statements (continued)
Theorem
If A is an n × n matrix and if TA : Rn → Rn is multiplication by A,
then the following are equivalent.
11. The row vectors of A span Rn .
12. The column vectors of A form a basis for Rn .
13. The row vectors of A form a basis for Rn .
14. rank(A) = n.
15. nullity(A) = 0.
16. The kernel of TA is {0}.
17. The range of TA is Rn .
18. TA is one-to-one.
19. λ = 0 is not an eigenvalue of A.
Linear Transformations
Definition
If V and W are vector spaces then the function T : V → W is
a linear transformation if for all u, v ∈ V and scalars c,
1. T (u + v) = T (u) + T (v).
2. T (c u) = c T (u).
If V = W , then T is called a linear operator on V .
Examples (1 of 3)
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If V and W are any two vector spaces the mapping
T : V → W defined as T (v) = 0 for all v ∈ V is called the
zero transformation.
Examples (1 of 3)
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If V and W are any two vector spaces the mapping
T : V → W defined as T (v) = 0 for all v ∈ V is called the
zero transformation.
T (u + v) = 0 = 0 + 0 = T (u) + T (v)
T (c u) = 0 = (c) 0 = c T (u)
Examples (1 of 3)
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If V and W are any two vector spaces the mapping
T : V → W defined as T (v) = 0 for all v ∈ V is called the
zero transformation.
T (u + v) = 0 = 0 + 0 = T (u) + T (v)
T (c u) = 0 = (c) 0 = c T (u)
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If V is any vector space the mapping T : V → V defined
as T (v) = v for all v ∈ V is called the
Examples (1 of 3)
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If V and W are any two vector spaces the mapping
T : V → W defined as T (v) = 0 for all v ∈ V is called the
zero transformation.
T (u + v) = 0 = 0 + 0 = T (u) + T (v)
T (c u) = 0 = (c) 0 = c T (u)
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If V is any vector space the mapping T : V → V defined
as T (v) = v for all v ∈ V is called the
T (u + v) = u + v = T (u) + T (v)
T (c u) = c u = c T (u)
Examples (2 of 3)
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If V is any vector space and k is a fixed scalar, the
mapping T : V → V defined as T (v) = k v is a linear
operator. If k > 1 this is called a dilation of V . If 0 < k < 1
this is called a contraction of V .
Examples (2 of 3)
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If V is any vector space and k is a fixed scalar, the
mapping T : V → V defined as T (v) = k v is a linear
operator. If k > 1 this is called a dilation of V . If 0 < k < 1
this is called a contraction of V .
T (u + v) = k (u + v) = k u + k v = T (u) + T (v)
T (c u) = k c u = c k u = c T (u)
Examples (2 of 3)
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If V is any vector space and k is a fixed scalar, the
mapping T : V → V defined as T (v) = k v is a linear
operator. If k > 1 this is called a dilation of V . If 0 < k < 1
this is called a contraction of V .
T (u + v) = k (u + v) = k u + k v = T (u) + T (v)
T (c u) = k c u = c k u = c T (u)
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The mapping T : Pn → Pn given by
T (p) = T (p(x)) = p(x − 1) is a linear operator.
Examples (2 of 3)
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If V is any vector space and k is a fixed scalar, the
mapping T : V → V defined as T (v) = k v is a linear
operator. If k > 1 this is called a dilation of V . If 0 < k < 1
this is called a contraction of V .
T (u + v) = k (u + v) = k u + k v = T (u) + T (v)
T (c u) = k c u = c k u = c T (u)
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The mapping T : Pn → Pn given by
T (p) = T (p(x)) = p(x − 1) is a linear operator.
T (p + q) = (p + q)(x − 1) = p(x − 1) + q(x − 1)
= T (p) + T (q)
T (c p) = (cp)(x − 1) = c p(x − 1) = c T (p)
Examples (3 of 3)
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The mapping D : C 1 (−∞, ∞) → C(−∞, ∞) defined by
D(f) = f 0 (x) is a linear transformation.
Examples (3 of 3)
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The mapping D : C 1 (−∞, ∞) → C(−∞, ∞) defined by
D(f) = f 0 (x) is a linear transformation.
D(f + g) = (f + g)0 (x) = f 0 (x) + g 0 (x) = D(f) + D(g)
D(c f) = (c f )0 (x) = c f 0 (x) = c D(f)
Examples (3 of 3)
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The mapping D : C 1 (−∞, ∞) → C(−∞, ∞) defined by
D(f) = f 0 (x) is a linear transformation.
D(f + g) = (f + g)0 (x) = f 0 (x) + g 0 (x) = D(f) + D(g)
D(c f) = (c f )0 (x) = c f 0 (x) = c D(f)
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The mapping
J : C(−∞, ∞) → C 1 (−∞, ∞) defined by
Z
x
J(f) =
f (t) dt is a linear transformation.
0
Z
J(f + g) =
x
Z
(f + g)(t) dt =
0
0
0
Z
t
f (t) dt +
0
= J(f) + D(g)
Z x
Z
J(c f) =
(c f )(x) = c
x
g(t) dt
0
x
f (t) dt = c J(f)
Eigenvalues of a Linear Operator
Definition
If T : V → V is a linear operator on a vector space V , then a
nonzero vector x ∈ V is called an eigenvector of T if T (x) is a
scalar multiple of x; that is,
T (x) = λ x
for some scalar λ. The scalar λ is called an eigenvalue of T ,
and x is said to be an eigenvector corresponding to λ.
Eigenvalues of a Linear Operator
Definition
If T : V → V is a linear operator on a vector space V , then a
nonzero vector x ∈ V is called an eigenvector of T if T (x) is a
scalar multiple of x; that is,
T (x) = λ x
for some scalar λ. The scalar λ is called an eigenvalue of T ,
and x is said to be an eigenvector corresponding to λ.
The subspace of all vectors in V for which T (x) = λ x is called
the eigenspace of T corresponding to λ.
Example
If D : C ∞ (−∞, ∞) → C ∞ (−∞, ∞) is the differentiation operator
and if λ is a constant, then
D(eλx ) = λeλx
so that λ is an eigenvalue of D and eλx is the corresponding
eigenvector.
Homework
I
Read Section 5.1
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Exercises: 1–25 odd