The image of the transpose of $A^T$ is the orthogonal complement of its kernel

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