09-240/Classnotes for Tuesday October 20

Definition

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

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=I_V

T∘S=I_W

T(O_V)=O_W

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...U_n) of V

α=(W₁...W_n) of W

by an earlier theorem, ∃ a l.t. T:V→W such that T(U_i)=W_i

(T(∑a_iu_i)=∑a_iT(u_i)=∑a_iu_i)

∃ a l.t. S:W→V s.t. S(W_i)=U_i

Claim

S∘T=I_v

T∘S=I_w

Proof

If u∈$\mathrm {V}$ unto U=∑a_iu_i

(S∘T)(u)=S(T(u))=S(T(∑a_iu_i))

=S(∑a_iw_i)=∑a_iu_i=u

⇒S∘T=I_v...

⇒Assume T&S as above exist

Choose a basis β= (U₁...U_n) of V

Claim

α=(W₁=Tu₁, W₂=Tu₂, ..., W_n=Tu_n)

is a basis of W, so dim W=n

Proof

α is lin. indep.

T(0)=0=∑a_iw_i=∑a_iTu_i=T(∑a_iu_i)

Apply S to both sides:

0=∑a_iu_i

So ∃_ia_i=0 as β is a basis

α Spans W

Given any w∈W let u=S(W)

As β is a basis find a_is in F s.t. v=∑a_iu_i

Apply T to both sides: T(S(W))=T(u)=T(∑a_iu_i)=∑a_iT(u_i)=∑a_iW_i ∴ I win!!! (QED)