Prove: $ \operatorname{Ker}(T)^\perp= \operatorname{Im}(T^*)$

Question

Let $T:V\to V$

Prove: $ \operatorname{Ker}(T)^\perp= \operatorname{Im}(T^*)$

If $v\in \operatorname{Im}(T^*)$ so $\exists w\in V:T^*w=v$ but how can I continue from here?

If $v\in \operatorname{Ker}(T)^\perp$ what does it say?

Answer 1

We have

\begin{align} x \in \ker T &\iff Tx = 0 \\ &\iff \langle Tx,y\rangle= 0, \forall y \in V \\ &\iff \langle x,T^*y\rangle= 0, \forall y \in V \\ &\iff x \perp \operatorname{Im} T^* \\ &\iff x \in (\operatorname{Im} T^*)^\perp \end{align}

so $\ker T = (\operatorname{Im} T^*)^\perp$. Now take the orthogonal complement.

Answer 2

Remark: the decomposition $V=A\oplus A^\perp$ only works in case $V$ is finite-dimensional.

First observe that if $b$ is an element of the kernel of $T$, then $\langle T^*a,b\rangle=\langle a,Tb\rangle=0$, thus $\textrm{Im}T^*\subset\textrm{Ker}T^{\perp}$. Similarly, $\textrm{Ker}T\subset{\textrm{Im}T^*}^{\perp}$. On the other hands, $V=\textrm{Ker}T\oplus\textrm{Ker}T^\perp=\textrm{Im}T^*\oplus{\textrm{Im}T^*}^{\perp}$. In order to prove the other inclusions, one just need (sorry for this jump, I'm not sure how to explain :D) to prove ${\textrm{Im}T^*}^{\perp}\cap\textrm{Ker}T^\perp=\{0\}$, which is true for if $x$ is an element of this intersection then $\langle T^*a,x\rangle=0$ for all vector $a$, hence $x$ is contained in the Kernel of $T$ and in its complement, so $x$ is $0$.