Let $(V,\langle,\rangle)$ be a finite dimensional vector space. Let $T$ be a linear operator on V. Show that $\operatorname{Im}T=(\ker T^*)^\perp$

Question

Help me proving this please. Let $(V,\langle,\rangle)$ be a finite dimensional vector space. Let $T$ be a linear operator on V. Show that $\operatorname{Im}T=(\ker T^*)^\perp$

Answer 1

Hints:

1. you have a finite dimensional inner product vector space
2. by definition, $\langle Tx,y\rangle=\langle x,T^*y\rangle$
3. $U^\perp=\{x\in V\mid \forall y\in U : \langle x,y\rangle=0\}$, where $U$ is a subspace of $V$
4. in a finite dimensional inner product space $V$, for every subspace $U$ of $V$ it holds $U^{\perp\perp}=U$

Because of the statements above, you need to prove that, for all $x\in\operatorname{Im}T$ and $y\in\operatorname{Ker}T^*$, then $\langle x,y\rangle=0$.

Now, supply the full argument.