4

Let $U$ and $V$ be vector spaces over a field $F$, and let $T: U\rightarrow V$ be a linear transformation, with $T^*: V^*\rightarrow U^*$ the corresponding adjoint. We would like to show that $\text{Im}(T^*)=\text{Ker}(T)^{\perp}$, where $S^{\perp}$ is defined to be $\{f\in U^*:f(S)=\{0\}\}$ for all subsets $S$ of $U$

I am able to show that the left hand side is contained in the right hand side, but not the other containment. How do we go about proving the other containment?

yoshi
  • 1,061
  • @DonAntonio It's the same definition given by OP –  May 09 '13 at 08:09
  • @yoshi Can you provide the source of this problem ? Because I don't think the other containment is true in general. Even if we assume $U,V$ to be topological vector sapces, you have $ ker(T)^\perp = \overline{Im(T^)}^ $ which is the weak* closure. You probably require $Im(T) $ to be closed furthermore. For finite dimensions it is fine. – smiley06 May 09 '13 at 08:22

1 Answers1

3

Let us write $N=\operatorname{ker}(T)$ and $W=\operatorname{im}(T)$.

Start with $f\in N^\perp$. We can define $g\in W^*$ by setting $g(Tx)=f(x)$ for all $x\in U$. This is well defined, for if $Tx=Ty$ then $x-y\in N$, so $f(x)=f(y)$.

Next, extend $g$ to a linear functional on $V$. This can be done directly by using Zorn's lemma, or we can pick a basis for $W$ and then extend it to $V$, then exted $g$ by setting $g(v)=0$ for each new basis vector.

So now $g\in V^*$ and $T^*g(x)=g(Tx)=f(x)$ for all $x\in U$, hence $T^*g=f$.

Harald Hanche-Olsen
  • 31,960