Lecture 31: Quick review from previous lecture

• A and AT has the same eigenvalues.

• tr(A) =Pn

i=1 aii =Pn

i=1 �i. (2) detA = �1�2 · · ·�n.

• If �1, . . . ,�k are pairwise distinct eigenvalues of A, then the correspondingeigenvectors v1, . . . ,vk are linearly independent.

• An eigenvalue � of a matrix A is complete if the number of linearly indepen-dent eigenvectors with eigenvalue � is equal to the multiplicity of �.

—————————————————————————————————Today we will discuss Diagonalization.

- Lecture will be recorded -

—————————————————————————————————

• Solutions for Midterm 2 has been posted on Canvas, see ”Announcements”.

MATH 4242-Week 13-1 1 Spring 2020

(def)- .-

e-

←

Example.

• A =

✓c 10 c

◆has only one eigenvalue c with multiplicity 2, with only one

eigenvector (1, 0)T . Its eigenvalue c is NOT a complete eigenvalue.

• On the other hand, the matrix B =

✓c 00 c

◆also has only one eigenvalue

c with multiplicity 2, but it had two linearly independent eigenvectors (1, 0)T

and (0, 1)T (and any non-zero linear combination of these). Its eigenvalue c isa complete eigenvalue.

Definition: If A is a matrix with eigenvalue �, we define the eigenspace of� to be

V� = ker(A� �I).

Then

dimV� = the number of linearly independent eigenvectors of A with eigenvalue �.

Thus, if dimV� = the multiplicity of �, then � is complete.*It can be shown that dimV� is never greater than �’s multiplicity.

Definition: If all eigenvalues of A are complete, we say the matrix A itself isa complete matrix.

Fact: If n ⇥ n matrix A is complete, then we can form a basis of Cn with itseigenvectors.

MATH 4242-Week 13-1 2 Spring 2020

o -

o-- det CA -Az) = det ( ton !. ) = K -a 5A -CI = ( I f) , Ker IA- ca ) = span / ( lol ) =VI. eigenspace

din Va = /

O = det (B- a 2) =CC- R 5 .

B-CI = ( 88 ),

Ken CB-CI) has basis ( (f),(9) f- Vc , eigenspace

with din -VEZ.

C)=

. A -- Ceo : ) is not complete• 13=1991 is complete .

I == =

§ Diagonalization. Consider the linear operator L[v] = Av. Suppose ma-trix A is complete, and v1, . . . ,vn are its basis of eigenvectors, with eigenvalues�1, . . . ,�n.

Q: What happens if we change basis, and represent A in the basis v1, . . . ,vn?

[To see this]. Denote by B the matrix that represents the operator L[v] = Av inthe basis v1, . . . ,vn. As we’ve seen,

B = V �1AV

where

V = [v1, . . . ,vn].

MATH 4242-Week 13-1 3 Spring 2020

- -

-

-

-11-I eigenvectors

4A.

V-

B = AV = A Cv. . . . ., on ]=CAr. . . .. . Arn]=LAri , . . .

,Turn]

If B = ( bi. . . . .bn ] , then H

V-

B = Vfb , . . . . .

,bn ] :-[Vb . . . .. .

Vbn).

Thus, Tfbj = Aj Vj , I E j E h .

Writing bj = f !! ) ,then

Ttb;= II bij Vi = Dj Vj

.

Since r. . . . . ✓n are linearly independent , thisimplies gbjyj.IO, #j .

It leads to 13=4! "-- non )#

From above, we have shown that we can factor any complete matrix A as fol-lows:

A = V DV �1

where V = [v1, . . . ,vn] is the matrix of eigenvectors, and

D = diag(�1, . . . ,�n) =

0

BBB@

�1 0 · · · 00 �2 · · · 0... . . . ...0 · · · 0 �n

1

CCCA

is the diagonal matrix of eigenvalues.

X In other words, representing the operator L[v] = Av in the basis of A’seigenvectors gives a diagonal matrix.

Definition: We say that the matrix A is diagonalizable, meaning it can befactored in the form A = V DV �1 where D is diagonal and V is nonsingular.

Fact: A matrix is complete if and only if it is diagonalizable.

MATH 4242-Week 13-1 4 Spring 2020

F-diag la , . - . .,Al = V-'AV

(B)

Let’s revisit the examples.

Example. A =

0

@3 1 01 3 00 0 2

1

A . We have found its eigenvalues � = 2, 2, 4.

Moreover,

eigenvalue � = 2, eigenvectors v1 = (�1, 1, 0)T , v2 = (0, 0, 1)T ,

eigenvalue � = 4, eigenvector v3 = (1, 1, 0)T .

Thus, the matrix A is complete. Moreover,

A = V DV �1,

where D = diag(2, 2, 4) and V = [v1,v2,v3].

Example. We already have found the eigenvectors and eigenvalues of the rotation

matrix Q✓ =

✓cos ✓ � sin ✓sin ✓ cos ✓

◆.

We’ve seen thatQ✓ has eigenvalues ei✓ and e�i✓, and eigenvectors (i, 1)T , (�i, 1)T .

MATH 4242-Week 13-1 5 Spring 2020

-Lecture

- complete

- complete

here: :# avi:T

EX : Tocomputev'

.

( in lecture 30 )

T. I T T

o. -- fi.-ille: Ii -it

'

= c : le:: t*

§ Some properties

• Suppose A = V DV �1, where D = diag(�1, . . . ,�n) and V = [v1, . . . ,vn].

What is A2 and Ak?

Fact: Ak has the same eigenvectors asA, and the eigenvalues are just �k1, . . . ,�

kn.

To generalized this fact:

• We say two matrices A and B are simultaneously diagonalizable if A =V D1V �1 and B = V D2V �1, for diagonal matrices D1 and D2.

MATH 4242-Week 13-1 6 Spring 2020

AZ = VDVV D-VI -V D2 DVIVDPV"

=-

U-D2-u-

T

=

In general ,Ak = Tt D

KV-t

.

= ar . . . oil ("

o

'

"

... cu - - - ri'

.

¢Lee Di -- (

"

o

'

:a:/ , R -- fo

'

:.non )-

A- #SB = V D

, V-'Tt Dz -V

-t= VD ,RV

"

-

=-

v ("

Y '

...

.v '

.

Also, BA = V f

"

?'

-

. ) V ".Thus

,AB = BA

. #

• If A = V DV �1, then A is invertible if and only if D = diag(�1, . . . ,�n) hasall nonzero diagonal elements.

Indeed,

detA =

So detA 6= 0 if and only if all �i 6= 0.

• Find A�1.

MATH 4242-Week 13-1 7 Spring 2020

e---

det ( VDV " )= deed - deed - dtv )= det D= 17,172 . - - An

.

A-' = ( VDV -15'

= fu - t)"

Dt Tf- I

= Tt D'

Tf-I

=-

o: u "

so, IT is eigenvalue of A"

.

are just the products of the eigenvalues of A and B.

Fact: If A and B are simultaneously diagonalizable. Then A and B have the same eigenvectors. And AB does too and the eigenvalues

§ Systems of Di↵erential Equations.Consider the system of di↵erential equations

x01 = 3x1 + x2 + x3x02 = 2x1 + 4x2 + 2x3x03 = �x1 � x2 + x3,

where xi = xi(t) is a di↵erentiable real-valued function of the real variable t.Clearly, xi(t) = 0 is the solution of the system.

— To find the general solutions to this system:

MATH 4242-Week 13-1 8 Spring 2020

-

e:÷H÷÷x¥,x'it, LA-

'

Xt's

We can diagonalize A into A -_ VDV?where D= ( I ) , -0=149 I1- A-2

, eigenvectors v, ,✓, Tk Ys.

"

17=4, s v, J

X't) = A xct , = VDV"X

=

⇒ VI ' =D -

V-'

x.

Let ytI=-V"xT .Then

.

y'

it , = D yet )

that isi÷ . all "% ):L: nmie.to solve .

'Y ,'

HK ⑦Yik);Litt C ,

{ Yz'HF 2 Yah, yah,= Cz e't ! 4,4, GEIR

,

y; HK 4 Y3H).

y3H1= Cs Ett.

so, xier-v-ycti-fyei.tl :÷÷.)

-- Eyal + ai://tea-ysf.tt)-✓ in Karla-42)

.inker (A - LI)#