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.
Best Answer
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, $\left<T^*v,w\right>=\left<v,Tw\right>=\left<v,0\right>=0$.