09-240/Classnotes for Tuesday October 20

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Definition

V & W are "isomorphic" if there exists a linear transformation T:V → W & S:W → V such that T∘S=IW and S∘T=IV


Theorem

If V& W are field dimensions over F, then V is isomorphic to W iff dim V=dim W


Corollary

If dim V = n then  \mathrm{V} \cong  \mathrm{F^n}

Note:  \cong  represents isomorphism

Two "mathematical structures" are "isomorphic" if there's a "bijection" between their elements which preserves all relevant relations between such elements.

Example: Plastic chess is "isomorphic" to ivory chess, but it is not isomorphic to checkers.

Ex: The game of 15. Players alternate drawing one card each. Goal: To have exactly three of your cards add to 15.

O: 7, 4, 6, 5 → Wins! X: 3, 8, 1, 2

This game is isomorphic to Tic Tac Toe!

4 9 2
3 5 7
8 1 6

Converts to:

O 9 X
X O O
X X O
S∘T=IV
T∘S=IW
T(OV)=OW
T(x+y)=T(x)+T(y)
T(cV)=cT(V)
Likewise for  \mathrm{S}
z=x+y ⇒ T(z)=T(x)+T(y)
u=7v ⇒ T(u)=7T(v)

Proof of Theorem  \Leftrightarrow Assume dim V= dim W=n

∃ basis β= (U1...Un) of V
α=(W1...Wn) of W
by an earlier theorem, ∃ a l.t. T:V→W such that T(Ui)=Wi

(T(∑aiui)=∑aiT(ui)=∑aiui)

∃ a l.t. S:W→V s.t. S(Wi)=Ui


Claim

S∘T=Iv
T∘S=Iw


Proof

If u∈ \mathrm{V} unto U=∑aiui

(S∘T)(u)=S(T(u))=S(T(∑aiui))
=S(∑aiwi)=∑aiui=u
⇒S∘T=Iv...
⇒Assume T&S as above exist
Choose a basis β= (U1...Un) of V

Claim

α=(W1=Tu1, W2=Tu2, ..., Wn=Tun)

is a basis of W, so dim W=n

Proof

α is lin. indep.

T(0)=0=∑aiwi=∑aiTui=T(∑aiui)
Apply S to both sides:
0=∑aiui
So ∃iai=0 as β is a basis

α Spans W

Given any w∈W let u=S(W)
As β is a basis find ais in F s.t. v=∑aiui

Apply T to both sides: T(S(W))=T(u)=T(∑aiui)=∑aiT(ui)=∑aiWi ∴ I win!!! (QED)