m211f2012 Slides Sec3.3handout

7/28/2019 m211f2012 Slides Sec3.3handout

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MATH 211 Winter 2013

Lecture Notes(Adapted by permission of K. Seyffarth)

Sections 3.3

Sections 3.3 Page 1/50


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3.3 Diagonalization and Eigenvalues



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Example

Let A = 4 2

1 3 . Find A100.

How can we do this efficiently?

Consider the matrix P = 1 21 1 . Observe that P is invertible (why?),

and that

P1 =1

3

1 2

1 1

.

Furthermore,

P1AP =1

3

1 2

1 1

4 2

1 3

1 21 1

=

2 00 5

= D,

where D is a diagonal matrix.



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Example (continued)

This is significant, because

P1AP = D

P(P1AP)P1 = PDP1

(PP1)A(PP1) = PDP1

IAI = PDP1

A = PDP1,and so

A100 = (PDP1)100

= (PDP1)(PDP1)(PDP1) (PDP1)

= PD(P1P)D(P1P)D(P1 P)DP1

= PDIDIDI IDP1

= PD100P1.



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Example (continued)

Now,

D100 =

2 00 5

100=

2100 00 5100

.

Therefore,

A100 = PD100P1

=

1 21 1

2100 0

0 5100

1

3

1 2

1 1

=1

3 2100 + 2 5100 2100 2 5100

2100 5100 2 2100 + 5100 =

1

3

2100 + 2 5100 2100 2 5100

2100 5100 2101 + 5100



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Theorem (3.3 Theorem 1)

If A = PDP1, then Ak = PDkP1 for each k = 1, 2, 3, . . .

The process of finding an invertible matrix P and a diagonal matrix D sothat A = PDP1 is referred to as diagonalizing the matrix A, and P is

called the diagonalizing matrix for A.

Problem

When is it possible to diagonalize a matrix?

How do we find a diagonalizing matrix?



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Eigenvalues and Eigenvectors

Definition

Let A be an n n matrix, a real number, and X = 0 an n-vector. IfAX = X, then is an eigenvalue of A, and X is an eigenvector of Acorresponding to , or a -eigenvector.

Example

Let A =

1 21 2

and X =

11

. Then

AX = 1 2

1 2

1

1 =

3

3 = 3

1

1 = 3X.

This means that 3 is an eigenvalue of A, and

11

is an eigenvector of A

corresponding to 3 (or a 3-eigenvector of A).



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Finding all Eigenvalues and Eigenvectors of a Matrix

Suppose that A is an n n matrix, X = 0 an n-vector, R, and that

AX = X.Then

X AX = 0

IX AX = 0

(I A)X = 0

Since X = 0, the matrix I A has no inverse, and thus

det(I A) = 0.



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Definition

The characteristic polynomial of an n n matrix A is

cA(x) = det(xI A).

Example

The characteristic polynomial of A = 4 21 3

iscA(x) = det

x 00 x

4 2

1 3

= det x 4 2

1 x 3

= (x 4)(x 3) 2

= x2 7x + 10



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Let A be an n n matrix.

1 The eigenvalues of A are the roots of cA(x).

2 The-eigenvectors X are the nontrivial solutions to (I A)X = 0.

Example (continued)

For A = 4 21 3

, we havecA(x) = x

2 7x + 10 = (x 2)(x 5),

so A has eigenvalues 1 = 2 and 2 = 5.

To find the 2-eigenvectors of A, solve (2I A)X = 0:2 2 0

1 1 0

1 1 0

2 2 0

1 1 00 0 0



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Example (continued)

The general solution, in parametric form, is

X =

t

t

= t

11

where t R.

To find the 5-eigenvectors of A, solve (5I A)X = 0:

1 2 01 2 0

1 2 00 0 0

The general solution, in parametric form, is

X =

2s

s

= s

2

1

where s R.



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Definition

A basic eigenvector of an n n matrix A is any nonzero multiple of a basicsolution to (I A)X = 0, where is an eigenvalue of A.

Example (continued)

11

and

21

are basic eigenvectors of the matrix

A =

4 2

1 3

corresponding to eigenvalues 1 = 2 and 2 = 5, respectively.



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Example

For A =

3 4 21 2 2

1 5 5

, find cA(x), the eigenvalues of A, and find

corresponding basic eigenvectors.

det(xI A) = x 3 4 2

1 x + 2 21 5 x 5

= x 3 4 2

0 x 3 x + 31 5 x 5

=

x 3 4 2

0 x 3 01 5 x

= (x 3) x 3 2

1 x

cA(x) = (x 3)(x2 3x + 2) = (x 3)(x 2)(x 1).



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Example (continued)

Therefore, the eigenvalues of A are 1 = 3, 2 = 2, and 3 = 1.

Basic eigenvectors corresponding to 1 = 3: solve (3I A)X = 0.

0 4 2 0

1 5 2 01 5 2 0

1 0 12 0

0 1 12 00 0 0 0

Thus X =

12 t1

2 t

t

= t

121

21

, t R.

Choosing t = 2 gives us X1 = 11

2

as a basic eigenvector correspondingto 1 = 3.



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Example (continued)

Basic eigenvectors corresponding to 2 = 2: solve (2I A)X = 0. 1 4 2 01 4 2 0

1 5 3 0

1 0 2 00 1 1 0

0 0 0 0

Thus X =

2ss

s

= s

21

1

, s R.

Choosing s = 1 gives us X2 = 2

11

as a basic eigenvector corresponding

to 2 = 2.



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Example (continued)

Basic eigenvectors corresponding to 3 = 1: solve (I A)X = 0. 2 4 2 01 3 2 0

1 5 4 0

1 0 1 00 1 1 0

0 0 0 0

Thus X =

rr

r

= r

11

1

, r R.

Choosing r = 1 gives us X3 = 1

11

as a basic eigenvector corresponding

to 3 = 1.



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Geometric Interpretation of Eigenvalues and EigenvectorsLet A be a 2 2 matrix. Then A can be interpreted as a lineartransformation from R2 to R2.

ProblemHow does the linear transformation affect the eigenvectors of the matrix?

Definition

Let V be a nonzero vector in R2. We denote by LV the unique line in R2that contains V and the origin.

Lemma (3.3 Lemma 1)

Let V =

ab

be a nonzero vector in R2. Then LV is the set of all scalar

multiples of V , i.e.,

LV = RV = {tV | t R} .



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Example (3.3 Example 6)

Let m R and consider the linear transformation Qm : R2 R2, i.e.,

reflection in the line y = mx.

The matrix that induces Qm is

A =1

1 + m2

1 m2 2m

2m m2 1

.

Claim. X1 =

1m

is an eigenvector of A corresponding to eigenvalue

= 1.

The reason for this: X1 = 1m lies in the line y = mx, and hence

Qm

1m

=

1m

, implying that A

1m

= 1

1m

.



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Example (continued)

More generally, any vector

k

km

, k = 0, lies in the line y = mx and is

an eigenvector of A.

Another way of saying this is that the line y = mx is A-invariant for thematrix

A =1

1 + m2

1 m2 2m

2m m2 1

.



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Example (3.3 Example 7)

Let be a real number, and R : R2 R2 rotation through an angle of,induced by the matrix

A = cos sin sin cos .

Claim. A has no real eigenvectors unless is an integer multiple of, i.e.,, 2, 3, etc.

The reason for this: a line L in R2 is A invariant if and only if is an

integer multiple of.


Di li ti


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Diagonalization

Notation. An n n diagonal matrix

D =

a1 0 0 0 00 a2 0 0 00 0 a3 0 0...

......

......

...

0 0 0 an1 00 0 0 0 an

is written diag(a1, a2, a3, . . . , an1, an).

Recall that if A is an n n matrix and P is an invertible n n matrix sothat P1AP is diagonal, then P is called a diagonalizing matrix of A, andA is diagonalizable.



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Let A be an n n matrix.1 A is diagonalizable if and only if it has eigenvectors X1,X2, . . . , Xn so

that

P =

X1 X2 Xn

is invertible.

2 If P is invertible, then

P1AP = diag(1, 2, . . . , n)

wherei is the eigenvalue of A corresponding to the eigenvector Xi,i.e., AXi = iXi.



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Note. It is not always possible to find n eigenvectors so that P isinvertible.

Example

Let A =

1 2 32 6 61 2 1

. Then

cA(x) =

x 1 2 32 x 6 61 2 x + 1

= = (x 2)3.

A has only one eigenvalue, 1 = 2, with multiplicity three.

To find the 2-eigenvectors of A, solve the system (2I A)X = 0.



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Example (continued)

1 2 3 02 4 6 01 2 3 0

1 2 3 00 0 0 00 0 0 0

The general solution in parametric form is

X =

2s + 3ts

t

= s

21

0

+ t

30

1

, s, t R.

Since the system has only two basic solutions, there are only two basiceigenvectors, implying that the matrix A is not diagonalizable.



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Example

Diagonalize, if possible, the matrix A =

1 0 10 1 0

0 0 3

.

cA(x) = det(xI A) =

x 1 0 10 x 1 00 0 x + 3

= (x 1)2(x + 3).

A has eigenvalues 1 = 1 of multiplicity two; 2 = 3 of multiplicity one.



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Example (continued)

Eigenvectors for 1 = 1: solve (I A)X = 0. 0 0 1 00 0 0 0

0 0 4 0

0 0 1 00 0 0 0

0 0 0 0

X =

st0

, s, t R so basic eigenvectors corresponding to 1 = 1 are

100 , 0

10



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Example (continued)

Eigenvectors for 2 = 3: solve (3I A)X = 0. 4 0 1 00 4 0 0

0 0 0 0

1 0 14 00 1 0 0

0 0 0 0

X = 14 t0

t

, t R so a basic eigenvector corresponding to 2 = 3 is

104



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Example (continued)

Let

P =

1 1 00 0 1

4 0 0

.

Then P is invertible, and

P1AP = diag(3, 1, 1) =

3 0 00 1 00 0 1

.



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A square matrix A is diagonalizable if and only if every eigenvalue of

multiplicity m yields exactly m basic eigenvectors, i.e., the solution to

(I A)X = 0 has m parameters.

A special case of this is


An n n matrix with distinct eigenvalues is diagonalizable.



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Example

Show that A =

1 1 00 1 0

0 0 2

is not diagonalizable.

First,

cA(x) =

x 1 1 00 x 1 00 0 x 2

= (x 1)2(x 2),so A has eigenvalues 1 = 1 of multiplicity two; 2 = 2 (of multiplicityone).



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Problem

Let A =

8 5 80 1 0

4 5 4

.

Show that4

is an eigenvalue of A, and find a corresponding basic

eigenvector.

Verify that

1 1 1

is an eigenvector os A, and find its

corresponding eigenvalue.



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Example


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Example

Consider the linear dynamical system Vk+1 = AVk with

A = 2 0

3 1, and V0 =

11

.

Find a formula for Vk.

First, cA(x) = (x 2)(x + 1), so A has eigenvalues 1 = 2 and 2 = 1,

and thus is diagonalizable.

Solve (2I A)X = 0:

0 0 03 3 0

1 1 00 0 0

has general solution X =

s

s

, s R, and basic solution X1 =

11

.



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Example (continued)

Solve (I A)X = 0:3 0 03 0 0

1 0 00 0 0

has general solution X = 0t, t R, and basic solution X2 = 0

1.

Thus, P =

1 01 1

is a diagonalizing matrix for A,

P

1

= 1 0

1 1, and P

1

AP = 2 0

0 1.



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Suppose thatVk = PD

kP1V0,

and assume that A has a dominant eigenvalue, 1, with corresponding

basic eigenvector X1 as the first column of P.For convenience, write P1V0 =

b1 b2 bn

T.


Th


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Then

Vk = PDkP1V0

=

X1 X2 Xn

k1 0 00 k2 0...

......

...0 0 kn

b1b2...

bn

= b1k1 X1 + b2k2 X2 + + bnknXn

= k1

b1X1 + b2

2

1

kX2 + + bn

n

1

kXn

Now, j1

< 1 for j = 2, 3, . . . n, and thus j1k 0 as k .Therefore, for large values of k, Vk

k1 b1X1.



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Example (continued)


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P1V0 =

1 0

1 1

1

1

=

1

2

=

b1b2

For large values of k,

Vk k1 b1X1 = 2

k(1) 11 = 2

k

2

k Lets compare this to the formula for Vk that we obtained earlier:

Vk = 2k2k 2(1)k

There is another example of estimating Vk for large values of k posted inthe Supplementary Notes section on Blackboard.


Example

Consider an owl population of 100 adult females and 40 juvenile females


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Consider an owl population of 100 adult females and 40 juvenile females,and assume we wish to study the population growth.

Imagine that we are biologists and have determined that in general:

1 The number of juvenile females hatched in any year is twice thenumber of adult females in the year before.

2 Half the adult females in any year survive to the next year.

3

One quarter of the juvenile females in any year survive to adulthood.Will the owls survive?

Let:

ak = the number of adult females after k yearsjk = the number of juvenile females after k years

Hence we get from the information above that:

ak+1 =

12 ak +

14jk

jk+1

= 2akSections 3.3 Page 45/50

Thus if we write Vk =

akj

, then we have:


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k

jk

,

V0 = 100

40 and

Vk+1 =

ak+1jk+1

=

12 ak + 14jk2ak

=

12

14

2 0

akjk

So we have the dynamical system:

Vk+1 = AVk where A =

12

14

2 0

, V0 =

100

40



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Calculations yield:

V0

= 10040

V1 =

8020

V2 = 70

160 V3 =

75

140

V4 =

72.5150

.....So what is your guess for the long term???


Lets se eigen al es!


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Let s use eigenvalues!

It turns out that A = 12

14

2 0 has (check it!):

Eigenvalue 1 = 1 and corresponding basic eigenvector X1 =

12

,

Eigenvalue 2 = 12 and corresponding basic eigenvector X2 =

1

4

Thus A is diagonalizable!

That is P1AP = D, where

P =

1 12 4

, P1 =

1

6

4 1

2 1

, and D =

1 00 12

.


Hence we get:


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akjk

= Vk = A

kV0

= PDkP

1V0

=

1 12 4

1 00 12

k16

4 1

2 1

10040

= 16 1 12 4

1k 00 ( 1

2)k

440160

= 16

1 ( 12 )

k

2 4( 12 )k

440

160

= 16

440 + 160( 12 )

k

880 640( 12 )k

Thus equating top and bottom entries yields exact formulas for akand jk:

ak =220

3+

80

3(

1

2)k and jk =

440

3

320

3(

1

2)k


So again the exact values are:

220 80 1 k 440 320 1 k


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ak =220

3+

80

3(

1

2)k and jk =

440

3

320

3(

1

2)k

For large values ofk:( 12 )

k is nearly 0, so we have the following approximate values:

ak 220

3and jk =

440

3

Hence, in the long term:

The female population stabilizes with approximately twice as manyjuveniles as adults. They survive!

Try the following:

What happens if the adult survival rate drops to 1/4?

What happens if the adult survival rate drops to 1/4, but the juvenilesurvival rate jumps to 1/2?

What ha ens if the adult survival rate dro s to 1 4 the uvenileSections 3.3 Page 50/50

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m211f2012 Slides Sec3.3handout