lecture 31: quick review from previous...
TRANSCRIPT
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)#