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)

T T

V → W ⇔ V' → W'

rank T=rank T'

Fix t:V→Wa l.t.$Insertformulahere$

Definition

1. N(T=ker(T)={u∈V:Tu=0_w}

2. R(T)=_im(T)={T(u):u∈V}

Prop/Def

1. N(T)'⊂V is a subspace of V-------nullity(T):=dim N(T)

2. R(T)⊂W is a subspace of W--------rank(T):=dim R(T)

Proof 1

x,y ∈N(T)⇒T(x)=0, T(y)=0

T(x+y)=T9x)+T(y)=0+0=0

x+y∈N(T)

∴ I win!!! (QED)

Proof 2

Let y∈R(T)⇒fix x s.t y=T(x),

--------7y=7T(x)=T(7x)

----------⇒7y∈R(T)

∴ I win!!! (QED)

Examples

1.

0:V→W---------N(0)=V

R(0)={0_W}-----------nullity(0)=dim V

--------------rank(0)=0

dim V+0=dimV

2.

I_V:V→V

N(I)={0}

nullity=0

R(I)=dim V

2'If T:V→W is an imorphism

N(T)={0}

nullity =0

R(T)=W

rank=dim W

0+dim V=dim V

3.

D:P₇(R)→P₇(R)

Df=f'

N(D)={C⊃C°: C∈R}=P₀(R)

R(D)⊂P₆(R)

nullity(D)=1

basis:(1x°)

rank(D)=7

7+1=8

4.

3':D²:P₇(R)

D²f=f

W(D²)={ax+b: a,b∈R}=P₁(R)

nullity(D²)=2

R(D²)=P₅(R)

rank (D²)=6

6+2=8

Theorem

(rank-nullity Theorem, a.k.a. dimension Theorem)

nullity(T)+rank(T)=dim V

(for a l.t. T:V→W) when V is F.d.

Proof

(To be continued next day)