The image of the transpose of $A^T$ 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 $$ImA^T=(KerA)^{\perp}$$ I know one direction: $ImA^T \subset (KerA)^{\perp}$ but I don't know how to show the other direction. Can anyone help me?

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 $x\notin \ker (A)^\perp$.

We need to show that $(Ker A)^\perp\subseteq ImA^T$. Taking orthogonal complements, this is equivalen tto $(ImA^T)^\perp\subseteq Ker A$.

Take $x\in (Im A^T)^\perp$. Since $V=Ker A\oplus(Ker A)^\perp$, we can write $x=y+z$, with $y\in Ker A$ and $z\in (Ker A)^\perp$. Then for all $v\in V$, \begin{align*} 0&=\langle x,A^Tv\rangle=\langle y,A^Tv\rangle+\langle z,A^Tv\rangle=\langle Ay,v\rangle+\langle Az,v\rangle=\langle Az,v\rangle \end{align*} where the last equality follows from $y\in Ker A$. Using the above with $v=Az$ yields $Az=0$, and so $z\in KerA\cap(KerA)^\perp=0$, so $z=0$ and $x=y\in KerA$.