09-240/Classnotes for Tuesday October 20: Difference between revisions

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Revision as of 09:49, 14 December 2009

WARNING: The notes below, written for students and by students, are provided "as is", with absolutely no warranty. They can not be assumed to be complete, correct, reliable or relevant. If you don't like them, don't read them. It is a bad idea to stop taking your own notes thinking that these notes can be a total replacement - there's nothing like one's own handwriting! Visit this pages' history tab to see who added what and when.


Definition: V and W are "isomorphic" if there exist linear transformations and such that and

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

Note: 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.

Goal: To have exactly three of your cards add to 15.

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
Likewise for

Proof of Theorem Assume dim(V) = dim(W) = n

There exists basis
by an earlier theorem, there exists a l.t. such that

There exists a l.t. such that


Claim


Proof

If u∈ 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.

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)

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