1

Math 2331

Section 5.1 – Eigenvectors and Eigenvalues

Example: Let 1 5

0 2A

====

2

1v

====

1

1w

==== −−−−

5

1u

====

Definition: An eigenvector of an n n×××× matrix A is a nonzero vector x such that

Ax xλλλλ==== for some scalar λλλλ .

A scalar λλλλ is an eigenvalue of A if there exists a nontrivial solution x of Ax xλλλλ==== ; such an x is called an eigenvector corresponding to λλλλ .

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Example: 4 2

1 3A

====

Is 2

1u

====

an eigenvector?

Is 1

4v

====

an eigenvector?

Example: 6 3

2 1A

====

Show that 7 is an eigenvalue of A.

3

Notice that Ax xλ=� �

for 0 0 ( ) 0λ λx Ax Ix A I x≠ ⇒ − = ⇒ − =�

.

That is, x�

is an eigenvector corresponding to λ if and only if ( )λx Nul A I∈ −�

.

Definition: If x�

is an eigenvector corresponding to λ , ( )Nul A Iλλλλ−−−−

is called the eigenspace for λλλλ .

Notation: Eλλλλ .

KEY FACT: λ is an eigenvalue of A if and only if det( ) 0λA I− = .

Reason:

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Definition: Suppose A is an nxn matrix.

The characteristic polynomial of A is an nth

degree polynomial in λ :

( ) det( )p A Iλ λ= −

Example: Let4 2

1 3A

=

. Find all e-values, e-vectors, e-spaces, characteristic

polynomial.

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FACT: The special solutions to ( ) 0A I xλ− = form a basis for Eλ . This basis consist

of e-vectors corresponding to e-value λ .

Example: Let1 1

1 1A

=

. Find all e-values, e-vectors, e-spaces, characteristic

polynomial.

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Example: Let0 1

1 0A

=

. Find all e-values, e-vectors, e-spaces, characteristic

polynomial.

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Example: Let

1 1 0

1 2 1

0 1 1

A

− = − − −

. Find all e-values, e-vectors, e-spaces,

characteristic polynomial.

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Example: Let

2 1 0

0 2 5

0 0 2

A

=

. Find all e-values, e-vectors, e-spaces, characteristic

polynomial.

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Example: Let

2 1 0

0 2 0

0 0 2

A

=

. Find all e-values, e-vectors, e-spaces, characteristic

polynomial.

Definition: Algebraic multiplicity of an eigenvalue λ is the degree of the

corresponding factor in the characteristic polynomial.

Geometric multiplicity of an eigenvalue is the dimension of the corresponding

eigenspace.

Fact: GM AM≤≤≤≤

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IMPORTANT FACTS:

1) An nxn matrix A has at most n distinct eigenvalues.

2) If A is a triangular matrix, then the eigenvalues of A are the diagonal entries.

2 1 1

0 5 4

0 0 6

A

− =

; 2, 5, 6.

3) The product of the eigenvalues of a matrix A equals det( )A .

4) The sum of the eigenvalues of a matrix A equals ( )trace A .

Here, 11 22( ) ... nntrace A a a a= + + + = the sum of the diagonal entries.

Example:

4 2

1 3A

=

Prdocut of the evalues: 10

Sum of the eigenvalues: 7 ; 2 ad 5.

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Example: A is a 2x2 matrix with determinant -20 and trace 1. What are the

eigenvalues of A?

5) 0 is an eigenvalue of A if and only if A is singular.

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6) If A is invertible and 0λ ≠ , then λ is an eigenvalue of A if and only if 1

λ is an

eigenvalue of 1A− . The evectors for λ and 1

λ are the same.

Note: Elimination changes eigenvalues!

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Theorem: If {{{{ }}}}1 2, , ..., rv v v are eigenvectors corresponding to distinct

eigenvalues 1 2, , ..., rλ λ λλ λ λλ λ λλ λ λ , then the set {{{{ }}}}1 2, , ..., rv v v is linearly independent.

Question: If x is an eigenvector of A, what is ?kA x ====

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Math 2331

Section 5.2 – The Characteristic Equation

:A n n×××× matrix

The characteristic equation of A : det( ) 0A Iλλλλ− =− =− =− =

A scalar λλλλ is an eigenvalue of A if and only if λλλλ satisfies the characteristic equation det( ) 0A Iλλλλ− =− =− =− = .

Characteristic polynomial of A : ( ) det( )p A Iλ λλ λλ λλ λ= −= −= −= −

Algebraic multiplicity of an eigenvalue λλλλ is its multiplicity as a root of ( )p λλλλ .

Example: The characteristic polynomial of a matrix is:

6 5 4( ) 9 10p λ λ λ λλ λ λ λλ λ λ λλ λ λ λ= − −= − −= − −= − −

Find the eigenvalues and their algebraic multiplicities.

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The Invertible Matrix Theorem (continued)

Let A be an n n×××× matrix. A is invertible if and only if

s. The number 0 is NOT an eigenvalue of A.

t. The determinant of A is not 0.

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Similarity

Let A and B be two n n×××× matrices.

A is similar to B if there exists an invertible matrix P such that 1P AP B−−−− ==== .

This is equivalent to saying that 1A PBP −−−−==== .

Similarity Transformation: 1A P AP−−−−֏

Theorem: If A and B are similar matrices, then they have the same characteristic polynomial and the same eigenvalues (with the same multiplicities).

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Similar matrices have the SAME:

• Eigenvalues • Determinant • Trace • Rank • Number of linearly independent eigenvectors • Jordan form (later)

DIFFERENT:

• 4 subspaces (row space, column space, etc.) • Eigenvectors

FACT: If 1B P AP−−−−==== and x��

is an eigenvector for A, then 1P x−−−−��

is an eigenvector for B.

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Warning: 2 1

0 2A

====

and 2 0

0 2B

====

have the same eigenvalues. However,

they are NOT similar!

Similar � same eigenvalues

Same eigenvalues � may be similar, may be not.

Example : 2 3

0 1A

====

and 4 1

0 5B

====

are not similar.

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Example : 2 3

0 1A

====

Find a matrix which is similar to A.

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Math 2331

Section 5.3 – Diagonalization

Recall: Diagonal matrices --

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Definition: :A n n×××× matrix

A can be diagonalized ( or “A is diagonalizable”) if A is similar to a DIAGONAL matrix D .

That is, if there exists an invertible matrix P and a diagonal matrix D such that 1A PDP −−−−==== .

Example: 2 3 1 1 2 0 1 1

0 5 0 1 0 5 0 1

−−−− ====

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Theorem: The Diagonalization Theorem

An n n×××× matrix A is diagonalizable if and only if A has n linearly independent eigenvectors.

In fact, 1A PDP −−−−==== , with D a diagonal matrix, if and only if the columns of P are n linearly independent vectors of A. In this case, the diagonal entries of D are eigenvalues of A that correspond, respectively, to the eigenvectors in P.

P: The eigenvector matrix for A

D: The eigenvalue matrix for A.

In other words, A is diagonalizable if it has enough eigenvectors to form a basis for n

ℝℝℝℝ . Such a basis is called an eigenvector basis of nℝℝℝℝ .

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Example: Diagonalize 4 2

1 3A

=

.

From Section 5.1: e-value 2, e-vector 1

1

−−−−

, e-value: 5, e-vector 2

1

5

Example: Diagonalize 1 1 0

1 2 1

0 1 1

A

− = − − −

From Section 5.1- e-values are 0, 1, 3.

e-vector for 0:

1

1

1

, e-vector for 1:

1

0

1

−−−−

, e-vector for 3:

1

2

1

−−−−

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FACTS:

1) A is diagonalizable if every eigenvalue has enough eigenvectors; GM=AM.

(That is, if the sum of the dimensions of eigenspaces equals n )

2) If an n n×××× matrix A has n distinct eigenvalues, then A is diagonalizable.

(Reason: the set of n corresponding e-vectors will be linearly independent.)

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Some matrices can not be diagonalized.

Example: 0 1

0 0

;

Example: 2 1 0

0 2 5

0 0 2

;

1

( 2 ) 0

0

N A I span

− =

8

(Extra) Example: Diagonalize 1 3 1

3 5 1

3 3 1

A

− − = − − −

3 2 2( ) 5 8 4 (1 )( 2)p λ λ λ λ λ λ= − + − + = − − ;

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Powers of A

If A is diagonalized as: 1A PDP −−−−==== , then

11 1

0

0

k

k k

kn

d

A PD P S S

d

− −

= =

⋱ .

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Similarly, for roots:

11/2 1/2 1 1

0

0 n

d

A PD P S S

d

− −

= =

⋱

FACT: If A is diagonalized as 1A PDP −−−−==== , then

11 1 1 1

1/

1/

0

0 n

d

A PD P P P

d

− − − −

= =

⋱.

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Example: Let 4 2

1 3A

=

. Find 4 1 1/2, ,A A A− .

We know: 14 2 1 2 2 0 1/ 3 2 / 3

1 3 1 1 0 5 1/ 3 1/ 3A PDP− − −

= = =

44 4 1

4

1 2 1/ 3 2 / 3 422 4062 0

1 1 1/ 3 1/ 3 203 2190 5A PD P− − −

= = =

1 1 1 1 2 1/ 2 0 1/ 3 2 / 3 3 /10 1/ 5

1 1 0 1/ 5 1/ 3 1/ 3 1/10 2 / 5A PD P− − − − − −

= = = −

1/2 1/2 1 1 2 2 0 1 2 2 2 5 2 2 2 51 1

1 1 1 13 30 5 2 5 2 2 5A PD P−

− − + − + = = = − + +

Question: What are the eigenvalues of 4A ?

What are the eigenvalues of kA ?

What are the eigenvalues of 6A ?

What are the eigenvalues of 1A− ?

What are the eigenvalues of 2A I+ ?

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Exercise: 4 3

2 1A

−−−− ==== −−−−

Compute 6A .

Example: True or False?

A:nxn , A is diagonalizable if it has n eigenvectors.

If A is diagonalizable, then it has n distinct eigenvalues.

If A is invertible, then it is diagonalizable.

A: 3x3, A has two eigenvalues, each eigenspace is one dimensional. Then A is diagonalizable.

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A: 5x5, A has two eigenvalues, one eigenspace is 3 dimensional, the other is 2 dimensional. Then A is diagonalizable.

A: 7x7, A has 3 eigenvalues, one eigenspace is 2 dimensional, another eigenspace is 3 dimensional. A is diagonalizable.

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Math 2331

Section 5.4 – Eigenvectors and Linear Transformations

Recall from 1.9: Any linear transformation : n mT →→→→ℝ ℝℝ ℝℝ ℝℝ ℝ can be implemented via left-multiplication by a matrix A.

What if :T V W→→→→ ?

The Matrix of a Linear Transformation

V: vector space, dim ,V n==== a basis: {{{{ }}}}1 2, , ..., nB b b b====

W: vector space, dim ,W m==== a basis: {{{{ }}}}1 2, , ..., mC c c c====

Let :T V W→→→→ be a linear transformation.

We can find a matrix M such that

[[[[ ]]]] [[[[ ]]]]( )C B

T x M x====

(The action of T on x can be viewed as left multiplication by M.)

The matrix representation of T (“matrix for T rela tive to the bases B and C”):

1 2( ) ( ) ( )nC C CM T b T b T b ====

⋯⋯⋯⋯

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Example: Let {{{{ }}}}1 2,B b b==== be a basis for V and {{{{ }}}}1 2 3, ,C c c c==== be a basis for W.

Let :T V W→→→→ be a linear transformation such that

1 1 2 3( ) 5T b c c c= − += − += − += − + , and 2 1 2 3( ) 2T b c c c= + −= + −= + −= + − .

Find the matrix M for T relative to the bases B and C.

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Case: If :T V V→→→→

M is called the matrix for T relative to B;

[[[[ ]]]] [[[[ ]]]]( )B B

T x M x====

Example: Let 2 2:T P P→→→→ be defined as ( ) 'dp

T p pdt

= == == == =

Let B be the standard basis for 2P .

a) Find the B-matrix for T.

b) For 2( ) 2 5p t t t= + += + += + += + + , ( ) ?T p ====

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Linear Transformations on nℝℝℝℝ :

Let A be the standard matrix for T.

If A is diagonalizable, then there is a basis B for nℝℝℝℝ consisting of eigenvectors of

A. In this case, the B-matrix for T is diagonal.

Theorem: Diagonal Matrix Representation

Suppose A is diagonalized as 1A PDP −−−−==== .

If B is the basis for nℝℝℝℝ formed from the columns of P, then D is the B-matrix for

the transformation :T x Ax→→→→ .

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Example: Define 2 2:T →→→→ℝ ℝℝ ℝℝ ℝℝ ℝ by ( )T x Ax==== where 4 2

1 3A

====

.

Find a basis B for 2ℝℝℝℝ such that the B-matrix for T is diagonal.

Recall from 5.3 that we diagonalized A as:

14 2 1 2 2 0 1/ 3 2 / 3

1 3 1 1 0 5 1/ 3 1/ 3A PDP− − −

= = =

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Fact: If : n nT →→→→ℝ ℝℝ ℝℝ ℝℝ ℝ is defined by ( )T x Ax==== and B is any basis for nℝℝℝℝ , then

the B-matrix for T is similar to A.

The set of all matrices similar to A coincides with the set of all matrix representations of the transformation x Ax֏֏֏֏ .

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Problems from Homework

Example: Let {{{{ }}}}1 2 3, ,E e e e==== be the standard basis for 3ℝℝℝℝ , let {{{{ }}}}1 2 3, ,B b b b==== be

a basis for a vector space V. Let 3:T V→→→→ℝℝℝℝ be a linear transformation such that

1 2 3 3 2 1 2 2 1 3 3( , , ) (2 ) (2 ) ( 3 )T x x x x x b x b x x b= − − + += − − + += − − + += − − + +

Find the matrix for T relative to the bases E and B.

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Example: Let 2 3:T P P→→→→ be defined as ( ( )) ( 3) ( )T p t t p t= += += += +

a) 2( ) 5 2p t t t= + += + += + += + + , ( ) ?T p ====

b) Find the matrix for T relative to the standard bases for 2P and 3P .

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Example: Let 32:T P →→→→ ℝℝℝℝ be defined as

( 1)

( ) (0)

(1)

p

T p p

p

−−−− ====

a) ( ) 2 4p t t= += += += + , ( ) ?T p ====

b) Find the matrix for T relative to the standard bases for 2P and 3ℝℝℝℝ .

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Example: Let 6 2

4 0A

− −− −− −− − ====

and 1 2

0 1,

1 2b b

−−−− = == == == =

, 1 2{ , }B b b====

Find the B-matrix for the transformation x Ax֏֏֏֏ .

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Important: Verify the statements in problems 19—23!

If A is invertible and similar to B, then B is invertible.

If A is similar to B, then 2A is similar to 2B .

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If B is similar to A and C is similar to A, then B is similar to C.

If A is diagonalizable and B is similar to A, then B is also diagonalizable.

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If 1B P AP−−−−==== and x is an eigenvector for A corresponding to an eigenvalue λλλλ ,

then 1P x−−−− is an eigenvector for B corresponding to λλλλ .

Math 2331

5.5 – Complex Eigenvalues

Example: Let0 1

1 0A

− =

. Find all e-values, e-vectors, e-spaces.

Complex Eigenvalues:

Let A be a matrix with real entries. If a biλ = + is an eigenvalue of A, then

a biλ = − is also an eigenvalue. Complex eigenvalues occur in conjugate pairs.

Any e-vector

1

2

3

x

x

x

for a biλ = + corresponds to an e-vector

1

2

3

x

x

x

for a biλ = − .

Real and Imaginary Parts of a vector:

For example, if

1 2

5

2

i

x i

i

++++ ==== −−−−

, then

1

Re( ) 0

2

x

====

and

2

Im( ) 5

1

x

==== −−−−

;

Re( ) Im( )x x x i= += += += + iiii