Post on 21-Jan-2022
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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.
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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: 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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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
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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
− =
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(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