# 09-240/Classnotes for Tuesday October 20

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Definition: V and W are "isomorphic" if there exist linear transformations $\mathrm{T : V \rightarrow W}$ and $\mathrm{S : W \rightarrow V}$ such that $\mathrm{T \circ S} = I_\mathrm{W}$ and $\mathrm{S \circ T} = I_\mathrm{V}$

Theorem: If V and W are finite-dimensional over F, then V is isomorphic to W iff dim(V) = dim(W)

Corollary: If dim(V) = n then $\mathrm{V} \cong F^n$

Note: $\cong$ represents "is isomorphic to"

Two "mathematical structures" are "isomorphic" if there exists 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.

Example: The game of 15. Players alternate drawing one card each.

Sample game:

• X picks 3
• O picks 7
• X picks 8
• O picks 4
• X picks 1
• O picks 6
• X picks 2
• O picks 5
• 4 + 6 + 5 = 15. O wins.

This game is isomorphic to Tic Tac Toe!

 4 9 2 3 5 7 8 1 6
X: 3, 8, 1, 2
O: 7, 4, 6, 5 -- Wins!

Converts to:

 O 9 X X O O X X O
$\mathrm{S \circ T} = I_\mathrm{V}$
$\mathrm{T \circ S} = I_\mathrm{W}$
$\mathrm T(O_\mathrm{V}) = O_\mathrm{W}$
$\mathrm T(x + y) = T(x) + T(y)$
$\mathrm T(cv) = c\mathrm T(v)$
Likewise for $\mathrm S$
$z = x + y \Rightarrow \mathrm T(z) = \mathrm T(x) + \mathrm T(y)$
$u = 7v \Rightarrow \mathrm T(u) = 7\mathrm T(v)$

Proof of Theorem $\iff$ Assume dim(V) = dim(W) = n

There exists basis $\beta = \{u_1, \ldots, u_n\} \in \mathrm V$
$\alpha = \{w_1, ..., w_n\} \in \mathrm W$
by an earlier theorem, there exists a l.t. $\mathrm{T : V \rightarrow W}$ such that $\mathrm T(u_i) = w_i$

$\mathrm T(\sum a_i u_i) = \sum a_i \mathrm T(u_i) = \sum a_i w_i$

There exists a l.t. $\mathrm{S : W \rightarrow V}$ such that $\mathrm S(w_i) = u_i$

## Claim

$\mathrm{S \circ T} = I_\mathrm{V}$
$\mathrm{T \circ S} = I_\mathrm{W}$

## 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)

T T
V → W ⇔ V' → W'
rank T=rank T'

Fix t:V→Wa l.t.$Insert formula here$

## Definition

1. N(T) = ker(T) = {u∈V : Tu = 0W}
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)={0W}-----------nullity(0)=dim V
--------------rank(0)=0
dim V+0=dimV

2.

IV: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)