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Similar Matrices and Diagonalization

Linear AlgebraMATH 2076

Section 5.3 Similarity & Diagonalization 24 March 2017 1 / 10

Similar Matrices and Diagonalizable Matrices

Two n× n matrices A and B are similar if and only if there is an invertiblematrix P such that A = PBP−1 (and then we also haveB = P−1AP = QAQ−1 where Q = P−1).

An n × n matrix A is diagonalizable if and only if it is similar to a diagonalmatrix; that is, there are a diagonal matrix D and an invertible matrix Psuch that A = PDP−1.

An n × n matrix A is diagonalizable if and only if there is an eigenbasisassoc’d with A; that is, there is a basis {~v1, ~v2, . . . , ~vn} for Rn such thateach vector ~vi is an eigenvector for A. When this holds, say withA~vi = λi~vi , we have

A = PDP−1 where P =[~v1 ~v2 . . . ~vn

]and D =

λ1 0 . . . 00 λ2 . . . 0...

...... 0

0 0 . . . λn

.



Two n× n matrices A and B are similar if and only if there is an invertiblematrix P such that A = PBP−1

(and then we also haveB = P−1AP = QAQ−1 where Q = P−1).



A = PDP−1 where P =[~v1 ~v2 . . . ~vn

]and D =

λ1 0 . . . 00 λ2 . . . 0...

...... 0

0 0 . . . λn

.






A = PDP−1 where P =[~v1 ~v2 . . . ~vn

]and D =

λ1 0 . . . 00 λ2 . . . 0...

...... 0

0 0 . . . λn

.




An n × n matrix A is diagonalizable if and only if it is similar to a diagonalmatrix; that is,

there are a diagonal matrix D and an invertible matrix Psuch that A = PDP−1.


A = PDP−1 where P =[~v1 ~v2 . . . ~vn

]and D =

λ1 0 . . . 00 λ2 . . . 0...

...... 0

0 0 . . . λn

.






A = PDP−1 where P =[~v1 ~v2 . . . ~vn

]and D =

λ1 0 . . . 00 λ2 . . . 0...

...... 0

0 0 . . . λn

.





An n × n matrix A is diagonalizable if and only if there is an eigenbasisassoc’d with A; that is,

there is a basis {~v1, ~v2, . . . , ~vn} for Rn such thateach vector ~vi is an eigenvector for A. When this holds, say withA~vi = λi~vi , we have

A = PDP−1 where P =[~v1 ~v2 . . . ~vn

]and D =

λ1 0 . . . 00 λ2 . . . 0...

...... 0

0 0 . . . λn

.





An n × n matrix A is diagonalizable if and only if there is an eigenbasisassoc’d with A; that is, there is a basis {~v1, ~v2, . . . , ~vn} for Rn such thateach vector ~vi is an eigenvector for A.

When this holds, say withA~vi = λi~vi , we have

A = PDP−1 where P =[~v1 ~v2 . . . ~vn

]and D =

λ1 0 . . . 00 λ2 . . . 0...

...... 0

0 0 . . . λn

.






A = PDP−1 where P =[~v1 ~v2 . . . ~vn

]and D =

λ1 0 . . . 00 λ2 . . . 0...

...... 0

0 0 . . . λn

.






A = PDP−1 where P =[~v1 ~v2 . . . ~vn

]and D =

λ1 0 . . . 00 λ2 . . . 0...

...... 0

0 0 . . . λn

.Section 5.3 Similarity & Diagonalization 24 March 2017 2 / 10

A 2× 2 Example

The matrix A =

[1 24 3

]has simple eigenvalues λ1 = 5 and λ2 = −1 with

assoc’d eigenvectors ~v1 =

[12

]and ~v2 =

[1−1

].

Therefore,

A = PDP−1 where P =

[1 12 −1

]and D =

[5 00 −1

].

But what does this mean??


A 3× 3 Example

The matrix A =

4 −1 0−1 5 −10 −1 4

has simple eigenvalues 3, 4, 6 with

associated eigenvectors

~v1 =

111

, ~v2 =−10

1

, ~v3 = 1−21

.Therefore,


1 −1 11 0 −21 1 1

and D =

3 0 00 4 00 0 6

.



A 3× 3 Example

The matrix A =

4 −1 0−1 5 −10 −1 4


associated eigenvectors ~v1 =

111

, ~v2 =−10

1

, ~v3 = 1−21

.

Therefore,


1 −1 11 0 −21 1 1

and D =

3 0 00 4 00 0 6

.



A 3× 3 Example

The matrix A =

4 −1 0−1 5 −10 −1 4


associated eigenvectors ~v1 =

111

, ~v2 =−10

1

, ~v3 = 1−21

.Therefore,


1 −1 11 0 −21 1 1

and D =

3 0 00 4 00 0 6

.



3× 3 Matrices with Simple and Double Eigenvalues

A2 =

1 4 40 3 20 2 3

has one simple eigenvalue 5 and one double eigenvalue 2

with associated eigenvectors ~v1 =

211

, ~v2 =100

, ~v3 = 0

1−1

.{~v1, ~v2, ~v3} is an eigenbasis assoc’d with A2, so A2 is diagonalizable.

A3 =

5 −6 01 −2 04 6 −1

has one simple eigenvalue 4 and one double

eigenvalue −1 with associated eigenvectors ~v1 =

616

, ~v2 =001

.There is no eigenbasis assoc’d with A3, so A3 is not diagonalizable.



A2 =

1 4 40 3 20 2 3


with associated eigenvectors

~v1 =

211

, ~v2 =100

, ~v3 = 0

1−1


A3 =

5 −6 01 −2 04 6 −1



616

, ~v2 =001




A2 =

1 4 40 3 20 2 3



211

, ~v2 =100

, ~v3 = 0

1−1

.

{~v1, ~v2, ~v3} is an eigenbasis assoc’d with A2, so A2 is diagonalizable.

A3 =

5 −6 01 −2 04 6 −1



616

, ~v2 =001




A2 =

1 4 40 3 20 2 3



211

, ~v2 =100

, ~v3 = 0

1−1

.{~v1, ~v2, ~v3} is an eigenbasis assoc’d with A2, so

A2 is diagonalizable.

A3 =

5 −6 01 −2 04 6 −1



616

, ~v2 =001




A2 =

1 4 40 3 20 2 3



211

, ~v2 =100

, ~v3 = 0

1−1


A3 =

5 −6 01 −2 04 6 −1



616

, ~v2 =001




A2 =

1 4 40 3 20 2 3



211

, ~v2 =100

, ~v3 = 0

1−1


A3 =

5 −6 01 −2 04 6 −1


eigenvalue −1 with associated eigenvectors

~v1 =

616

, ~v2 =001




A2 =

1 4 40 3 20 2 3



211

, ~v2 =100

, ~v3 = 0

1−1


A3 =

5 −6 01 −2 04 6 −1



616

, ~v2 =001

.

There is no eigenbasis assoc’d with A3, so A3 is not diagonalizable.



A2 =

1 4 40 3 20 2 3



211

, ~v2 =100

, ~v3 = 0

1−1


A3 =

5 −6 01 −2 04 6 −1



616

, ~v2 =001

.There is no eigenbasis assoc’d with A3, so

A3 is not diagonalizable.



A2 =

1 4 40 3 20 2 3



211

, ~v2 =100

, ~v3 = 0

1−1


A3 =

5 −6 01 −2 04 6 −1



616

, ~v2 =001



Recall that A =

[1 24 3


assoc’d eigenvectors

~v1 =

[12

]and ~v2 =

[1−1

]. Since A’s two eigenvectors

are LI, they form an eigenbasis B = {~v1, ~v2}.

Suppose ~x = c1~v1 + c2~v2; so[~x]B =

[c1c2

].

Look atA~x =

A(c1~v1 + c2~v2

)=

c1A~v1 + c2A~v2 =

5c1~v1 − c2~v2

which says that

[A~x]B=

[5c1−c2

]=

[5 00 −1

] [c1c2

]=

[5 00 −1

] [~x]B.

Thus using B-coordinates, the action of A is just multiplication by thediagonal matrix

D =

[5 00 −1

]=

[λ1 00 λ2

].

But, how do we get A~x?

Recall that A =

[1 24 3



[12

]and ~v2 =

[1−1

].

Since A’s two eigenvectors



[c1c2

].

Look atA~x =

A(c1~v1 + c2~v2

)=

c1A~v1 + c2A~v2 =

5c1~v1 − c2~v2

which says that

[A~x]B=

[5c1−c2

]=

[5 00 −1

] [c1c2

]=

[5 00 −1

] [~x]B.


D =

[5 00 −1

]=

[λ1 00 λ2

].


Recall that A =

[1 24 3



[12

]and ~v2 =

[1−1




[c1c2

].

Look atA~x =

A(c1~v1 + c2~v2

)=

c1A~v1 + c2A~v2 =

5c1~v1 − c2~v2

which says that

[A~x]B=

[5c1−c2

]=

[5 00 −1

] [c1c2

]=

[5 00 −1

] [~x]B.


D =

[5 00 −1

]=

[λ1 00 λ2

].


Recall that A =

[1 24 3



[12

]and ~v2 =

[1−1



Suppose ~x = c1~v1 + c2~v2; so

[~x]B =

[c1c2

].

Look atA~x =

A(c1~v1 + c2~v2

)=

c1A~v1 + c2A~v2 =

5c1~v1 − c2~v2

which says that

[A~x]B=

[5c1−c2

]=

[5 00 −1

] [c1c2

]=

[5 00 −1

] [~x]B.


D =

[5 00 −1

]=

[λ1 00 λ2

].


Recall that A =

[1 24 3



[12

]and ~v2 =

[1−1




[c1c2

].

Look atA~x =

A(c1~v1 + c2~v2

)=

c1A~v1 + c2A~v2 =

5c1~v1 − c2~v2

which says that

[A~x]B=

[5c1−c2

]=

[5 00 −1

] [c1c2

]=

[5 00 −1

] [~x]B.


D =

[5 00 −1

]=

[λ1 00 λ2

].


Recall that A =

[1 24 3



[12

]and ~v2 =

[1−1




[c1c2

].

Look atA~x = A

(c1~v1 + c2~v2

)=

c1A~v1 + c2A~v2 =

5c1~v1 − c2~v2

which says that

[A~x]B=

[5c1−c2

]=

[5 00 −1

] [c1c2

]=

[5 00 −1

] [~x]B.


D =

[5 00 −1

]=

[λ1 00 λ2

].


Recall that A =

[1 24 3



[12

]and ~v2 =

[1−1




[c1c2

].

Look atA~x = A

(c1~v1 + c2~v2

)= c1A~v1 + c2A~v2 =

5c1~v1 − c2~v2

which says that

[A~x]B=

[5c1−c2

]=

[5 00 −1

] [c1c2

]=

[5 00 −1

] [~x]B.


D =

[5 00 −1

]=

[λ1 00 λ2

].


Recall that A =

[1 24 3



[12

]and ~v2 =

[1−1




[c1c2

].

Look atA~x = A

(c1~v1 + c2~v2

)= c1A~v1 + c2A~v2 = 5c1~v1 − c2~v2

which says that

[A~x]B=

[5c1−c2

]=

[5 00 −1

] [c1c2

]=

[5 00 −1

] [~x]B.


D =

[5 00 −1

]=

[λ1 00 λ2

].


Recall that A =

[1 24 3



[12

]and ~v2 =

[1−1




[c1c2

].

Look atA~x = A

(c1~v1 + c2~v2

)= c1A~v1 + c2A~v2 = 5c1~v1 − c2~v2

which says that[A~x]B=

[5c1−c2

]=

[5 00 −1

] [c1c2

]=

[5 00 −1

] [~x]B.


D =

[5 00 −1

]=

[λ1 00 λ2

].


Let’s draw a picture for the coordinate mapping ~x 7→[~x]B, and also for

the inverse map too.

~x plane ~c plane

~c =[~x]B

~x = P~c

~v1

~v2

B = {~v1, ~v2}

~x = P~c ~c =[~x]B

Here ~c =[~x]B =

[c1c2

]means ~x = c1~v1 + c2~v2 =

[~v1 ~v2

]~c = P

[~x]B; so the

B to S change of coordinates matrix P is given by

P = PSB =

[~v1 ~v2

]=

[1 12 −1

].



~x plane ~c plane

~c =[~x]B

~x = P~c

~v1

~v2

B = {~v1, ~v2}

~x = P~c ~c =[~x]B

Here ~c =[~x]B =

[c1c2

]means

~x = c1~v1 + c2~v2 =[~v1 ~v2

]~c = P

[~x]B; so the


P = PSB =

[~v1 ~v2

]=

[1 12 −1

].



~x plane ~c plane

~c =[~x]B

~x = P~c

~v1

~v2

B = {~v1, ~v2}

~x = P~c ~c =[~x]B

Here ~c =[~x]B =

[c1c2

]means ~x = c1~v1 + c2~v2 =

[~v1 ~v2

]~c = P

[~x]B; so

the


P = PSB =

[~v1 ~v2

]=

[1 12 −1

].



~x plane ~c plane

~c =[~x]B

~x = P~c

~v1

~v2

B = {~v1, ~v2}

~x = P~c ~c =[~x]B

Here ~c =[~x]B =

[c1c2

]means ~x = c1~v1 + c2~v2 =

[~v1 ~v2

]~c = P

[~x]B; so the


P = PSB =

[~v1 ~v2

]=

[1 12 −1

].



~x plane ~c plane

~c =[~x]B

~x = P~c

~v1

~v2

B = {~v1, ~v2}

~x = P~c ~c =[~x]B

Here ~c =[~x]B =

[c1c2

]means ~x = c1~v1 + c2~v2 =

[~v1 ~v2

]~c = P

[~x]B; so the


P = PSB =[~v1 ~v2

]=

[1 12 −1

].

From an earlier slide: WTF A~x and we know[A~x]B= D

[~x]B where D =

[5 00 −1

]and A =

[1 24 3

]and we have an eigenbasis assoc’d with A given by

B = {~v1, ~v2} where ~v1 =

[12

], ~v2 =

[1−1

].

Recall that for any vector ~w in R2 we have~w = P

[~w]B and

[~w]B = P−1~w

where the B to S change of coordinates matrix P is given by

P = PSB =

[~v1 ~v2

]=

[1 12 −1

].

Thus [~x]B = P−1~x and

A~x =

P[A~x]B=

PD[~x]B =

PDP−1~x .

So,

A = PDP−1 where D =

[5 00 −1

]=

[λ1 00 λ2

], P =

[~v1 ~v2

]=

[1 12 −1

].

Thus A and D are similar matrices. How to “see” the LT ~x 7→ A~x?


[~x]B where D =

[5 00 −1

]and A =

[1 24 3


B = {~v1, ~v2} where ~v1 =

[12

], ~v2 =

[1−1

].

Recall that for any vector ~w in R2 we have

~w = P[~w]B and

[~w]B = P−1~w


P = PSB =

[~v1 ~v2

]=

[1 12 −1

].


A~x =

P[A~x]B=

PD[~x]B =

PDP−1~x .

So,


[5 00 −1

]=

[λ1 00 λ2

], P =

[~v1 ~v2

]=

[1 12 −1

].



[~x]B where D =

[5 00 −1

]and A =

[1 24 3


B = {~v1, ~v2} where ~v1 =

[12

], ~v2 =

[1−1

].


[~w]B and

[~w]B = P−1~w


P = PSB =

[~v1 ~v2

]=

[1 12 −1

].


A~x =

P[A~x]B=

PD[~x]B =

PDP−1~x .

So,


[5 00 −1

]=

[λ1 00 λ2

], P =

[~v1 ~v2

]=

[1 12 −1

].



[~x]B where D =

[5 00 −1

]and A =

[1 24 3


B = {~v1, ~v2} where ~v1 =

[12

], ~v2 =

[1−1

].


[~w]B and

[~w]B = P−1~w


P = PSB =[~v1 ~v2

]=

[1 12 −1

].


A~x =

P[A~x]B=

PD[~x]B =

PDP−1~x .

So,


[5 00 −1

]=

[λ1 00 λ2

], P =

[~v1 ~v2

]=

[1 12 −1

].



[~x]B where D =

[5 00 −1

]and A =

[1 24 3


B = {~v1, ~v2} where ~v1 =

[12

], ~v2 =

[1−1

].


[~w]B and

[~w]B = P−1~w


P = PSB =[~v1 ~v2

]=

[1 12 −1

].

Thus [~x]B = P−1~x and A~x =

P[A~x]B=

PD[~x]B =

PDP−1~x .

So,


[5 00 −1

]=

[λ1 00 λ2

], P =

[~v1 ~v2

]=

[1 12 −1

].



[~x]B where D =

[5 00 −1

]and A =

[1 24 3


B = {~v1, ~v2} where ~v1 =

[12

], ~v2 =

[1−1

].


[~w]B and

[~w]B = P−1~w


P = PSB =[~v1 ~v2

]=

[1 12 −1

].

Thus [~x]B = P−1~x and A~x = P

[A~x]B=

PD[~x]B =

PDP−1~x .

So,


[5 00 −1

]=

[λ1 00 λ2

], P =

[~v1 ~v2

]=

[1 12 −1

].



[~x]B where D =

[5 00 −1

]and A =

[1 24 3


B = {~v1, ~v2} where ~v1 =

[12

], ~v2 =

[1−1

].


[~w]B and

[~w]B = P−1~w


P = PSB =[~v1 ~v2

]=

[1 12 −1

].


[A~x]B= PD

[~x]B =

PDP−1~x .

So,


[5 00 −1

]=

[λ1 00 λ2

], P =

[~v1 ~v2

]=

[1 12 −1

].



[~x]B where D =

[5 00 −1

]and A =

[1 24 3


B = {~v1, ~v2} where ~v1 =

[12

], ~v2 =

[1−1

].


[~w]B and

[~w]B = P−1~w


P = PSB =[~v1 ~v2

]=

[1 12 −1

].


[A~x]B= PD

[~x]B = PDP−1~x .

So,


[5 00 −1

]=

[λ1 00 λ2

], P =

[~v1 ~v2

]=

[1 12 −1

].



[~x]B where D =

[5 00 −1

]and A =

[1 24 3


B = {~v1, ~v2} where ~v1 =

[12

], ~v2 =

[1−1

].


[~w]B and

[~w]B = P−1~w


P = PSB =[~v1 ~v2

]=

[1 12 −1

].


[A~x]B= PD

[~x]B = PDP−1~x .

So,


[5 00 −1

]=

[λ1 00 λ2

], P =

[~v1 ~v2

]=

[1 12 −1

].

Thus A and D are similar matrices.

How to “see” the LT ~x 7→ A~x?


[~x]B where D =

[5 00 −1

]and A =

[1 24 3


B = {~v1, ~v2} where ~v1 =

[12

], ~v2 =

[1−1

].


[~w]B and

[~w]B = P−1~w


P = PSB =[~v1 ~v2

]=

[1 12 −1

].


[A~x]B= PD

[~x]B = PDP−1~x .

So,


[5 00 −1

]=

[λ1 00 λ2

], P =

[~v1 ~v2

]=

[1 12 −1

].


~c plane

~x plane

~d plane

~y plane

~y = A~x

A =

[1 24 3

]

~d = D~c

D =

[5 00 −1

]

~c =[~x]B ~x = P~c ~d =

[~y]B~y = P ~d

~v1

~v2

Diagonalizable Matrices

An n × n matrix A is diagonalizable if and only if there is an eigenbasisassoc’d with A. This holds if, say, A has n distinct (real) eigenvalues,because then the assoc’d eigenvalues are LI and hence form a basis.

In general, there is an eigenbasis assoc’d with A if and only if thedimensions of the eigenspaces for A add up to n.

Suppose λ is an eigenvalue for A. This means that λ is a zero for thecharacteristic polynomial pA of A. Therefore, we can factor pA as

p(t) = (t − λ)mq(t) for some m.

We call m the algebraic multiplicity of the eigenvalue λ.

We always have 1 ≤ dim E(λ) ≤ m.

We call dim E(λ) the geometric multiplicity of the eigenvalue λ.

There is an eigenbasis assoc’d with A if and only if for every eigenvalue λthe geometric multiplicity of λ equals the algebraic multiplicity of λ.



An n × n matrix A is diagonalizable if and only if there is an eigenbasisassoc’d with A.

This holds if, say, A has n distinct (real) eigenvalues,because then the assoc’d eigenvalues are LI and hence form a basis.










An n × n matrix A is diagonalizable if and only if there is an eigenbasisassoc’d with A. This holds if, say, A has n distinct (real) eigenvalues,because

then the assoc’d eigenvalues are LI and hence form a basis.












Suppose λ is an eigenvalue for A. This means that

λ is a zero for thecharacteristic polynomial pA of A. Therefore, we can factor pA as










Suppose λ is an eigenvalue for A. This means that λ is a zero for thecharacteristic polynomial pA of A.

Therefore, we can factor pA as















There is an eigenbasis assoc’d with A if and only if for every eigenvalue λ

the geometric multiplicity of λ equals the algebraic multiplicity of λ.


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