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Finite Sets Finite - dim v. spaces . If × mn ' If ( x ) = fqmalwlineqq.in bos × Y ""¥ ' F' " I :L :p" " * * one . Eth - FAI x them . got P g. of " Compo s . of linear maps off , y is the linen map of Compo of " Functor from to VectorSp categories r

Sets Finite - University of Toronto

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Finite Sets Finite - dim v. spaces . If

×mn '

If (x ) = fqmalwlineqq.inbos

× Y ""¥' F' "

I:L :p.

""

**one.

Eth- FAIx them.

gotPg.of

"

Compo s . of linear maps off , yis the linen map of Compo of "

Functor

from to VectorSp← categories r

f.d.

Review : V,W v.split ⇒ LIV

,w) 7 A can be described as

a dimwxdimv matrix of elts in # : choose bases

p fnv ,8 for W and write

ma .nn=*e=l::÷÷÷÷÷÷:)Changing fi , 8 will affect the matrix kxk kxn ath

to fir'

⇒ jfa) ,-

-

g. ftp.CAIpplITp '- -

fuk invertible nminvertible

Q =p. p

matrix matrix.

P Q-1

To emphasize the fact that the initial and final kxn

matrices represent the same linear map ,we can define

A,A'

two matrices to be "

equivalent"

when I P,Q invertible

matrices . sit '

A' = PA

Implements edoiumu operations ,invertible#kxk

('

off . - -toI' it

Simplify given kxu matrix A without changing its

equivalence class.

① GE (Row ops) to RRE

( o⇐Ei¥¥E¥¥② GE (Col . ops ) to clear all unknowns (* )

rent

f③ Column switching to bring everything to tight .

I offset diagonal.Thnx:

Every kxn matrixAis equiv .

to

O ←rxr

block matrix #-) r --REA

O O

In terms of bases : for any linear

map A : V→W

F bases f" I sit

.

( e ,. - - ten)

"

( fi,. .. ,f←) .

A- ( e.) = f ,Ale?)= fz

:-

Acer) ='

fr

Ace ;) = o ti > r

rlA7p=ffo)

Recall studied maps X Is X⇒self - hips of a set .

f-. got

a nun,'

l- 2 • 3 q4 represent themap f . in cycle

g y notation (234①-fixed

point cycle

There is a corresponding description for Linear operators"

Linear operators on V"

= L ( V ,V ) .

Key difference : since domain t codomain coincide,

-

we need not choose two bases ;

one basis for V suffices .

i. e.

AE LCV ,V ) has a matrix after singlechoice p = ( e . , -- - , en) basis forV .

Hide -

-

{ It:]matrix .

⇒ have a different equivalence relation on uxu matrices

Def Two nxn matrices A,A'

are similar or conjugatewhen I P invertible man sit

.

-

A'

= PAP"

Motivation if we change f to f'

then

fats ' ' . etat , LIF--

p r> p"

the fact that column ops are forced by the row ops[ means that simplification does not occur by GE . ]⇒ new approach is needed to

"

understand"

the operator: spectral theoryeigenvector analysis

What kind of operators are there?

-

LN ,V) .

ypole =L!! ! ! ingdep .

o) O : v 1-30 .

fr .

basin.

!

1) I : ri- v tr- effs = Igi!

2) ZEE XI : v te Xv

" A. ↳ *← omaha #

a'""tiffs?)

--

,

written in Ad basis1

I i:S", n.

. I µ=ce,p(A)

p ¥> Ae .

K e

. -

-

re, r

lr

( r

-- -

-

While this operator is simple ( easy to understand when written

as above,if we change basia

,won't be simple i

⇒ ⇐÷: Eri;÷µ¥E÷;¥?=":e: same operator !

p'( Ifs PCA ]pp( I ]p , different basis !

E- ( 1¥.

),l-Y://eei.ee) .

In the above example ,e, ,e, play the role of

"fixed pts"

e.

'

,e: net

e., er are examples of Eigenvectors .

the scale factors 1,72 areexamples of

Eigenvalues

4) besides eigenvectors, there is another

type of behaviour :

cydes-e.atto 'detone eigenreet .

⇐% , ez 1-7 e ,

ez 1-1 ez

B -- Ce . . - - ien) bani for ✓

B'-

- Cei,. . .

,en

'

) e! c- V't

--

- i

e : ( en -- I eikjl=loi=j①(ed = 0

:

Pep ( en) -- o

B'islia.indepylet#a±fit-oWTS ai =o fi .

Strategy: show a,=o

,then show 92=0

,~ . -

r

apply LHS to e.

: a,5taze t. .-9= 9

,

RHS to e,

: ocqy = o.

}⇒a

apply LHS to ez

aa,It are

Az

RHS to ee = o,

}arepeat . . .

.D

%% ¥itsapply LHS te e

,

e.'

(ed tea'Ce) = Ito=L

.

RHS =O1=0 # .

Polynomials : x't 1 =/ ( x -7.) (x - Iz)extend 11 / /numbers

real(x - illxti )

once the extension IRC ¢ is done,

we can factorall polynomials

plz) = aot a,Z t 9272 t - - - t an 't"

ai E Cl

= an ( z - H) (z-W -

-a 17 - Xu)

XXe

-

IdF Te Liv,V)

Def: Invariant subspace UE V linear subspace

sit . TCU) E U i.e. Tu E U tf u E U .

✓ °

,

"

i.Tv

,

given TELCV, y

1¥inn

.

Ex⑥ to } E V int .

TreoAreNull

T① Null T

② RangeLT) i.e.

I v EV u-

-Tr

u

U

Tu = TuTu C- Range (Tl .

Deff: an eigenvalue of T is a number TEF

sit.F V nonzero with reNahi

'Tv = .lv/T-aIyIo

"⇒

The set of eigenvalues in tf is called the SPECTRUMnotice this means Spanks) is an Invariant suhspaeeoft

Warning.

.

..

I -:

T OElf is

a valid eigenval .

Any veto in Null (T ) is an eigenvectorwith eigenvalue O

.

Howtofindeigenveotors.su/veAv--Tv

i.e . (A - II ) v = O

2- step procedure-

TO find value I sit.

A - XI

has nullity > oi.e

.A -II not injective

i. e. A - II not surjective .

i.e.rank CA -7 I) c drinkto :÷:÷:÷÷÷::÷: "

( note Null A = So }

Egri A- = ( 28¥ ! ) so o net eigenvalue .)

①A - II '

- f-

{ I, u!) Q: awaken Italian space.

I do GEt

- 21, 2 o

O -T l

27-51714-7702-57114712

02714-71-50217111471-512° °[email protected]:( III)( o →

.

! .ir! ¥o o ingYmthis will have nonzero Null space ⇐ 1=1 or

7=2.⇒ 1=1

,1=2 are

②the a eigenvalues for mi operator . !¥7

choose 1=1 first- plugin explicit value .

Three (Existence of an eigenvalue over F- Q )

PI suppose din V= n choose

any nonzero

vector VE V .

MUST

(v,Tv

,Thr

,Thr

,. - - .

,Tnr) ← BE

UN DEP! ! !

i.e. F ai not allzero sit .

Aorta,Tv t - -

- - t ant"

v I 0

(AOI ta,T t a

,T't - - - t anTn) v = O

#plz) .

view this as a polynomial !with cuffs in field Cl !

= an (T- RI) (T- rat) --

-

- (T- rat ).

v

whoots of plz). in ¢ .

product of ft- r,I) has nonzero

hull space

i.e.

not invertible

⇒ atleast one of the futons is not invertible ,

i. e. (T- ri) has a nonzero null space

⇒ eighth D, =Fi

.

eigenvector is ? ? find it.

-

if we can find a based of

eigenvectorsB : ( e , > . . . , en)

thene

-

- fig ?;%÷ :?)this finding Diagonalof a basis of

Matrix.

eighredfycalled.piag.inalizarin" .

The

V, ,Vz eik with eigenvalues41 Xz and 71¥12 .

Then ( v , ,vz) tin. indep -

Pfi suppose a ,v , t am=D

then applyT a.Tv , t qtr , = To-

- O

-②

aid ,r , tazkrz=o

Want to show 9=0 and az=0

to show : take 210 - ②

a. Hair , =o⇒am

-

fo to

a : tube 7,0 -② A

Az ( X ,-7274=0

( 69 ]=A Ae.- e

,

A- ez -_ ez

single eigenvalue 2nd space ofeigenvectors .

If ;) Here .

→- -3

ed .-3 -

←¥i.

-

1." it ÷÷÷: :en te en

t en - ,

A- 'I -- Loo's g) I .

ey→ e3

-- l

up µ¥¥Tf1µ - space of states = { up , down } -- S

-- -11C ( s) = { a , - Cup) t a. (down) :

a' laid }down I

e"

basisIasi.

¥9"""ist:÷.

ef II up

In Quantum mechanics,

the light switch

of ( up)-

- 1 has Quantum state = a vector

of ( downy. -1 in ECS)

operator :"

Position"

operator Q-

÷:÷÷:ii÷: !"down ll c l l

le ( i c- Li

att ) -- ⑦ to )classicalstheap

← ystts

t¥*Ei

Riemann ① classicalsphere N Leighan"BYdfn" states