Proving $(\ker{T})^{\perp}\subseteq \operatorname{Im} T^{*}$ Let $V$ be a finite inner product space with $T\colon V\to V$  a linear transformation.
How can I prove that,   $(\ker{T})^{\perp}\subseteq \operatorname{Im}T^{*}$ ?
Edit:
My purpose is to prove that:
$\operatorname{rank}(T)=\operatorname{rank}(T^{*})$

I thought of an alternative proof to what you originally wanted to show:
If $T = [A]$ in an orthonormal basis (exists because we're in a finite space) $[T^*] = \bar{A}^t = A^*$.
$rank A^* = rank \left(\bar{A}^t \right)$ and because the conjugate does not change the rank(tell me if you need a proof to this one as well) it's the same as:
$rank \left(A^t\right)$ but as you've probably proved that $\forall M_{n\times n} rank_C\left(M \right) = rank_R\left(M \right) $ so the transpose does not have any change the rank so it equals to $rank \left(A\right)$.
Which therefore means that  given the isomorphism of matrices and the linear transformation they describe that $\operatorname{rank}(T)=\operatorname{rank}(T^{*})$. 
