Adjoint of transformation proof

[zcd]

Homework Statement

Let V be an inner product space and let T be a linear operator on V. Prove the following results:
a)R(T^*)^⊥=N(T)
b)R(T^*)=N(T)^⊥ if V is finite dimensional

Homework Equations

<Tx,y>=<x,T^*y>

The Attempt at a Solution

pick x and y ≠0 and <Tx,y>=<x,T^*y>=0
this implies x∈N(T) and x ⊥ R(T^*) , so N(T)⊇R(T^*)^⊥

This half of the subset relationship I got, but how do I prove the other way? And as for part b), do I just prove that (W^⊥)^⊥=W? If so, why did they mention that V is finite dimensional?

 

[lanedance]

The Attempt at a Solution

pick x and y ≠0 and <Tx,y>=<x,T^*y>=0
this implies x∈N(T) and x ⊥ R(T^*) , so N(T)⊇R(T^*)^⊥

i don't think this implies x∈N(T) and x ⊥ R(T^*)...

but that (x∈N(T) OR Tx ⊥ y) AND x ⊥ R(T^*)

 

[lanedance]

how about starting by either assuming you have vector in the nullspace, and then for the other direction a vector perpindicular to the row vectors of T*

 

[zcd]

With your suggestion, I tried:

pick x and y ≠0 (the 0 case is readily proven) and x∈N(T)
then <Tx,y>=<x,T*y>=0, which implies x ⊥ R(T*) i.e x∈R(T*)^⊥
N(T)⊆R(T*)^⊥

for the other way, pick x∈R(T*)^⊥ (x and y ≠0 again)
<Tx,y>=<x,T*y>=0 only when Tx=0, so x∈N(T)
N(T)⊇R(T*)^⊥

is that a sufficient proof?

 

[lanedance]

first part pretty much, however I would write it as follows (some of it probably just diffenrent style)
-->
first pick x∈N(T)
then Tx=0
and <Tx,y>=0 for all y
but <Tx,y>=<x,T*y>=0 for all y
T*y represents a linear combination of the columns of T*
x ⊥column space of T*
x ⊥ R(T*)
so x∈R(T*)^⊥
-----

actually now that I'm working thorugh it, what exactly do you mean by R & N?

 

[zcd]

R is the range of transformation T and N is the nullspace of transformation T. I don't think we're supposed to use columns of T as that implies we pick a basis to represent T as a matrix.

 

[lanedance]

ok well in the last post as y is arbitrary, you have shown x is perp to any element in the image of T (which depending on terminology can be the same as the range)