09240/Classnotes for Tuesday October 20

Definition: V and W are "isomorphic" if there exist linear transformations and such that and
Theorem: If V and W are finitedimensional 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
Contents 
Claim
Proof
If u∈ unto U=∑a_{i}u_{i}
 (S∘T)(u)=S(T(u))=S(T(∑a_{i}u_{i}))
 =S(∑a_{i}w_{i})=∑a_{i}u_{i}=u
 ⇒S∘T=I_{v}...
 ⇒Assume T&S as above exist
 Choose a basis β= (U_{1}...U_{n}) of V
Claim
α=(W_{1}=Tu_{1}, W_{2}=Tu_{2}, ..., W_{n}=Tu_{n})
 is a basis of W, so dim W=n
Proof
α is lin. indep.
 T(0)=0=∑a_{i}w_{i}=∑a_{i}Tu_{i}=T(∑a_{i}u_{i})
 Apply S to both sides:
 0=∑a_{i}u_{i}
 So ∃_{i}a_{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_{i}s in F s.t. v=∑a_{i}u_{i}
Apply T to both sides: T(S(W))=T(u)=T(∑a_{i}u_{i})=∑a_{i}T(u_{i})=∑a_{i}W_{i}
 ∴ I win!!! (QED)
 T T
 V → W ⇔ V' → W'
 rank T=rank T'
Fix t:V→Wa l.t.
Definition
 N(T) = ker(T) = {u∈V : Tu = 0_{W}}
 R(T) = _{i}m(T) = {T(u) : u∈V}
Prop/Def
 N(T) ⊂ V is a subspace of Vnullity(T) := dim N(T)
 R(T) ⊂ W is a subspace of Wrank(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→WN(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:P_{7}(R)→P_{7}(R)
 Df=f'
 N(D)={C⊃C°: C∈R}=P_{0}(R)
 R(D)⊂P_{6}(R)
 nullity(D)=1
 basis:(1x°)
 rank(D)=7
 7+1=8
4.
 3':D^{2}:P_{7}(R)
 D^{2}f=f
 W(D^{2})={ax+b: a,b∈R}=P_{1}(R)
 nullity(D^{2})=2
 R(D^{2})=P_{5}(R)
 rank (D^{2})=6
 6+2=8
Theorem
(ranknullity 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)