Kernel and image of a matrix and its transposition

Two standard facts.

(i) $\langle Ax,y\rangle=\sum_i\big(\sum_ja_{ij}x_j\big)y_i=\sum_jx_j\big(\sum_ia_{ij}y_i\big)=\langle x,A^Ty\rangle$.

(ii) For a subspace $W\subseteq\mathbb{R}^k$: extending an orthonormal basis of $W$ shows $\mathbb{R}^k=W\oplus W^\perp$, so $\dim W^\perp=k-\dim W$. Also $W\subseteq(W^\perp)^\perp$, and both have dimension $\dim W$, hence $(W^\perp)^\perp=W$.

First identity. For $x\in\mathbb{R}^n$:
$$x\in\ker A\iff Ax=0\iff\langle Ax,y\rangle=0\ \forall y\in\mathbb{R}^m\iff\langle x,A^Ty\rangle=0\ \forall y\in\mathbb{R}^m.$$
The last condition says that $x$ is orthogonal to every vector $A^Ty$, i.e. to $\operatorname{Im} A^T$. Hence $\ker A=(\operatorname{Im} A^T)^\perp$.

Second identity. Apply the first identity to $A^T$ (whose transpose is $A$): $\ker A^T=(\operatorname{Im} A)^\perp$. Taking orthogonal complements and using (ii): $(\ker A^T)^\perp=\operatorname{Im} A$.

Rank. By (ii), $\dim\ker A=\dim(\operatorname{Im} A^T)^\perp=n-\operatorname{rank}A^T$. By rank-nullity, $\dim\ker A=n-\operatorname{rank}A$. Hence $\operatorname{rank}A=\operatorname{rank}A^T$ (row rank equals column rank).

Conclusion. Both identities hold, and they imply $\operatorname{rank}A=\operatorname{rank}A^T$.