The image of the transpose of $A^\text{T}$ is the orthogonal complement of its kernel Suppose $V$ is a finite dimensional vector space over $\mathbb{K}$. Let $A$ be a linear map. I am trying to prove that $$\operatorname{Im}A^\text{T}=(\operatorname{Ker}A)^{\perp}$$
I know one direction: $\operatorname{Im}A^\text{T} \subset (\operatorname{Ker}A)^{\perp}$ but I don’t know how to show the other direction. Can anyone help me?
 A: If $x\notin{\rm im}(A^T)$, note that ${\rm im}(A^T)={\rm im}(A^T)^{\perp\perp}$ because $V$ is finite dimensional. So there exists 
$x'\in{\rm im}(A^T)^\perp$ such that $ \langle x,x'\rangle\ne0$.
In fact, we have $x'\in\ker (A)$ because $A^TAx'\in{\rm im}(A^T)$, which implies
$$\langle Ax', Ax'\rangle=\langle x',A^TAx'\rangle=0\quad\Longrightarrow\quad
Ax'={\it 0}.$$
Thus $x\notin \ker (A)^\perp$.
A: We need to show that $(\operatorname{Ker} A)^\perp\subseteq \operatorname{Im} A^\text{T}$. Taking orthogonal complements, this is equivalent to $(\operatorname{Im} A^\text{T})^\perp\subseteq \operatorname{Ker} A$.
Take $x\in (\operatorname{Im} A^\text{T})^\perp$. Since $V= \operatorname{Ker} A\oplus( \operatorname{Ker} A)^\perp$, we can write $x=y+z$, with $y\in \operatorname{Ker} A$ and $z\in ( \operatorname{Ker} A)^\perp$. Then for all $v\in V$,
\begin{align*}
0&=\langle x,A^\text{T}v\rangle\\&=\langle y,A^\text{T}v\rangle+\langle z,A^\text{T}v\rangle\\&=\langle Ay,v\rangle+\langle Az,v\rangle\\&=\langle Az,v\rangle
\end{align*}
where the last equality follows from $y\in \operatorname{Ker} A$. Using the above with $v=Az$ yields $Az=0$, and so $z\in \operatorname{Ker}A\cap( \operatorname{Ker}A)^\perp=0$, so $z=0$ and $x=y\in \operatorname{Ker} A$.
