TITLE: The range of $T^*$ is the orthogonal complement of $\ker(T)$
QUESTION [21 upvotes]: 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.

REPLY [10 votes]: 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 $x\in\ker(T)$.