Lesson 15: Diagonalization

Lesson 15 (S&H, Section 14.5)Diagonalization

Math 20

October 24, 2007

Announcements

I Midterm done. Nice job!

I Problem Set 6 assigned today. Due October 31.

I OH: Mondays 1–2, Tuesdays 3–4, Wednesdays 1–3 (SC 323)

I Prob. Sess.: Sundays 6–7 (SC B-10), Tuesdays 1–2 (SC 116)

Math 20 - October 24, 2007.GWBWednesday, Oct 24, 2007

Page1of13

Outline

Concept Review

DiagonalizationMotivating ExampleProcedure

Computations

Uses of Diagonalization

Concept Review

DefinitionLet A be an n × n matrix. The number λ is called an eigenvalueof A if there exists a nonzero vector x ∈ Rn such that

Ax = λx. (1)

Every nonzero vector satisfying (1) is called an eigenvector of Aassociated with the eigenvalue λ.

Geometric effect of a non-diagonal linear transformation

Example

Let A =

(1/2 3/23/2 1/2

). Draw the effect of the linear transformation

which is multiplication by A.

x

y

v1

Av1

v2

v2

2e2

Ae2


Example

Let A =

(1/2 3/23/2 1/2



x

y

v1

Av1

v2

v2

2e2

Ae2


Example

Let A =

(1/2 3/23/2 1/2



x

y

v1

Av1

v2

v2

2e2

Ae2


Example

Let A =

(1/2 3/23/2 1/2



x

y

v1

Av1

v2

v2

2e2

Ae2


Example

Let A =

(1/2 3/23/2 1/2



x

y

v1

Av1

v2

v2

2e2

Ae2


Example

Let A =

(1/2 3/23/2 1/2



x

y

v1

Av1

v2

v2

2e2

Ae2


Example

Let A =

(1/2 3/23/2 1/2



x

y

v1

Av1

v2

v2

2e2

Ae2


Example

Let A =

(1/2 3/23/2 1/2



x

y

v1

Av1

v2

v2

2e2

Ae2


Example

Let A =

(1/2 3/23/2 1/2



x

y

v1

Av1

v2

v2

2e2

Ae2


Example

Let A =

(1/2 3/23/2 1/2



x

y

v1

Av1

v2

v2

2e2

Ae2


Example

Let A =

(1/2 3/23/2 1/2



x

y

v1

Av1

v2

v2

2e2

Ae2


Example

Let A =

(1/2 3/23/2 1/2



x

y

v1

Av1

v2

v2

2e2

Ae2

Methods

I To find the eigenvalues of a matrix A, find the determinant ofA− λI. This will be a polynomial in λ (called thecharacteristic polynomial of A, and its roots are theeigenvalues.

I To find the eigenvector(s) of a matrix corresponding to aneigenvalue λ, do Gaussian Elimination on A− λI.

Outline

Concept Review


Computations


Math 20 - October 24, 2007.GWBWednesday, Oct 24, 2007

Page5of13

The fact that we have eigenvectors corresponding to two differenteigenvalues gives us the following:

A

(1 11 −1

)︸︷︷︸

P

=

(A

(11

)A

(1−1

))=

(2

(11

)−1

(1−1

))

=

(1 11 −1

)(2 00 −1

)︸︷︷︸

D

So we have found a matrix P and a diagonal matrix D such that

AP = PD

Since P is invertible (det P = −2), we have

A = PDP−1


A

(1 11 −1

)︸︷︷︸

P

=

(A

(11

)A

(1−1

))=

(2

(11

)−1

(1−1

))

=

(1 11 −1

)(2 00 −1

)︸︷︷︸

D


AP = PD


A = PDP−1


A

(1 11 −1

)︸︷︷︸

P

=

(A

(11

)A

(1−1

))=

(2

(11

)−1

(1−1

))

=

(1 11 −1

)(2 00 −1

)︸︷︷︸

D


AP = PD


A = PDP−1

Diagonalization Procedure

I Find the eigenvalues and eigenvectors.

I Arrange the eigenvectors in a matrix P and the correspondingeigenvalues in a diagonal matrix D.

I If you have “enough” eigenvectors so that the matrix P issquare and invertible, the original matrix is diagonalizable andequal to PDP−1.

Outline

Concept Review


Computations


Example

Example (Worksheet Problem 1)

Let

A =

(0 −2−3 1

).

Find an invertible matrix P and a diagonal matrix D such thatA = PDP−1.

SolutionWe found that −2 and 3 are the eigenvalues for A. The eigenvalue

−2 has an associated eigenvector

(11

), and the eigenvalue 3 has

eigenvector

(−23

). Thus

P =

(1 −21 3

)D =

(−2 00 3

).

Example


Let

A =

(0 −2−3 1

).

Find an invertible matrix P and a diagonal matrix D such thatA = PDP−1.

SolutionWe found that −2 and 3 are the eigenvalues for A. The eigenvalue

−2 has an associated eigenvector

(11

), and the eigenvalue 3 has

eigenvector

(−23

). Thus

P =

(1 −21 3

)D =

(−2 00 3

).

Checking the Solution

P =

(1 −21 3

)D =

(−2 00 3

).

Check this: We have

P−1 =1

5

(3 2−1 1

).

PDP−1 =1

5

(1 −21 3

)(−2 00 3

)(3 2−1 1

)=

1

5

(1 −21 3

)(−6 −4−3 3

)=

1

5

(0 −10−15 5

)=

(0 −2−3 1

).


Let B =

(−7 4−9 5

). Find an invertible matrix P and a diagonal

matrix D such that B = PDP−1.

SolutionThe characteristic polynomial of B is (λ+ 1)2, which has the

double root −1. There is one eigenvector,

(23

), but nothing more.

So there is no diagonal D which works.


Let B =

(−7 4−9 5



SolutionThe characteristic polynomial of B is (λ+ 1)2, which has the

double root −1. There is one eigenvector,

(23

), but nothing more.

So there is no diagonal D which works.


Let B =

(0 1−1 0



SolutionThe characteristic polynomial of B is λ2 + 1, which has no realroots. The eigenvalues are i and −i . We could consider thecomplex eigenvectors

z1 =

(i1

)and z2 =

(−i1

)but scaling by a complex number is more complicated than it looks.


Let B =

(0 1−1 0



SolutionThe characteristic polynomial of B is λ2 + 1, which has no realroots. The eigenvalues are i and −i . We could consider thecomplex eigenvectors

z1 =

(i1

)and z2 =

(−i1

)but scaling by a complex number is more complicated than it looks.

Outline

Concept Review


Computations


Geometric effects of powers of matrices

Example

Let D =

(2 00 −1


which is multiplication by Dn.

x

y

S

D(S)

D2(S)

D3(S)


Example

Let D =

(2 00 −1



x

y

S

D(S)

D2(S)

D3(S)


Example

Let D =

(2 00 −1



x

y

S

D(S)

D2(S)

D3(S)


Example

Let D =

(2 00 −1



x

y

S

D(S)

D2(S)

D3(S)


Example

Let A =

(1/2 3/23/2 1/2


which is multiplication by An.

x

y

S

A(S)

A2(S)


Example

Let A =

(1/2 3/23/2 1/2



x

y

S

A(S)

A2(S)


Example

Let A =

(1/2 3/23/2 1/2



x

y

S

A(S)

A2(S)


Example

Let A =

(1/2 3/23/2 1/2



x

y

S

A(S)

A2(S)

Computing An with diagonalization


Let A =

(1/2 3/23/2 1/2

). Find A100.

Solution

We know A = PDP−1, where P =

(1 11 −1

)and D =

(2 00 −1

).

Now

An = (PDP−1)n = (PDP−1)(PDP−1) · · · (PDP−1)︸︷︷︸n

= PD(P−1P)D(P−1P) · · ·D(P−1P)DP−1 = PDnP−1

And Dn is easy! So

A100 =1

2

(1 11 −1

)(2100 0

0 1

)(1 11 −1

)=

1

2

(2100 + 1 2100 − 12100 − 1 2100 + 1

)



Let A =

(1/2 3/23/2 1/2

). Find A100.

Solution


(1 11 −1

)and D =

(2 00 −1

).

Now



And Dn is easy! So

A100 =1

2

(1 11 −1

)(2100 0

0 1

)(1 11 −1

)=

1

2

(2100 + 1 2100 − 12100 − 1 2100 + 1

)



Let A =

(1/2 3/23/2 1/2

). Find A100.

Solution


(1 11 −1

)and D =

(2 00 −1

).

Now



And Dn is easy!

So

A100 =1

2

(1 11 −1

)(2100 0

0 1

)(1 11 −1

)=

1

2

(2100 + 1 2100 − 12100 − 1 2100 + 1

)



Let A =

(1/2 3/23/2 1/2

). Find A100.

Solution


(1 11 −1

)and D =

(2 00 −1

).

Now



And Dn is easy! So

A100 =1

2

(1 11 −1

)(2100 0

0 1

)(1 11 −1

)=

1

2

(2100 + 1 2100 − 12100 − 1 2100 + 1

)

Technology

Lesson 15: Diagonalization