Kernel of the transpose

Equivalence. Let $y\in\mathbb{R}^m$. Then
$$y\in\ker(A^T)\iff A^T y=0\iff \text{for all }x\in\mathbb{R}^n,\ (A^T y)\cdot x=0.$$
But $(A^T y)\cdot x = y\cdot(Ax)$, so this is equivalent to $y\cdot(Ax)=0$ for all $x$, i.e. $y\perp\operatorname{Im} A$.

Conclusion. $\ker(A^T)=(\operatorname{Im} A)^\perp$.

Consequence. $\operatorname{rank}(A^T)=m-\dim\ker(A^T)=m-\dim(\operatorname{Im} A)^\perp=m-(m-\operatorname{rank}A)=\operatorname{rank}A$. So row rank equals column rank.