The range of $T^*$ is the orthogonal complement of $\ker(T)$

How can I prove that, if $V$ is a finite-dimensional vector space with inner product and $T$ a linear operator in $V$, then the range of $T^*$ is the orthogonal complement of the null space of $T$?

I know what I must do (for a $v$ in the range of $T^*$, I have to show that $v\perp w$ for every $w$ in $\ker(T)$ and then do the opposite), but I don't know how to show that this inner product is zero.

In order to show that the range of $T^*$ is the orthogonal complement of $\ker T$, we have to show that $\forall v \in \operatorname{Im}T^*$, $\forall w\in \ker T$: $\left<v,w\right>=0$.

Note that vectors in the range of $T^*$ are of the form $T^*v$ for $v\in V$. Now, let $w\in\ker T $. We have to show that $\left<T^*v,w\right>=0$. And, indeed, $\left<T^*v,w\right>=\left<v,Tw\right>=\left<v,0\right>=0$.

Let $A=\operatorname{ran}(T^*), B=\ker(T)^\perp$.

$\boxed{A\subseteq B:}$

For $x\in A, \ x=T^*y\ $ for some $y\in V$. Then, for any $z\in \ker(T),\ \langle x,z\rangle=\langle T^*y,z\rangle=\langle y,Tz\rangle=\langle y,0\rangle=0.$ Hence $x\in B.$

$\\ \\ \boxed{B\subseteq A:}$

Because $V$ is finite dimensional and $A,B$ is subspace, it is equivalent to $A^\perp \subseteq B^\perp= \ker(T)$. If $x\in A^\perp$, for any $y\in V$, $0=\langle x,T^*y\rangle=\langle Tx,y\rangle$ Therofore, $Tx=0$ (see exercise 8.1.1 (b) or simply take $y=Tx$) , and thus $x\in\ker(T)$.