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.

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

O: 7, 4, 6, 5 -- Wins!

X: 3, 8, 1, 2

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=∑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)