= Solution
Two <matrices> are equivalent when $A'=PAQ$ for invertible <matrices> $P$ and $Q$ of the appropriate sizes. A change of <basis> in the codomain multiplies a <matrix> on the left, while a change of <basis> in the domain multiplies it on the right. Hence two <matrices> of the same <linear map> in two pairs of <bases> are equivalent.
Conversely, start with the map $\alpha:F^n\to F^m$ having <matrix> $A$ in the standard <bases>. Given $A'=PAQ$, choose domain and codomain <bases> whose change-of-coordinate <matrices> produce $Q$ and $P$; then $A'$ represents the same map in those <bases>. This proves the equivalence.
The column rank of $A$ is the dimension of the span of its columns, and its row rank is the dimension of the span of its rows. If $A$ represents $\alpha:F^n\to F^m$, its column rank is $\operatorname{rank}\alpha$. The transpose represents the dual map
$$
\alpha^*:(F^m)^*\to(F^n)^*,
\qquad \phi\mapsto\phi\circ\alpha,
$$
so the row rank is $\operatorname{rank}\alpha^*$.
If $\operatorname{rank}\alpha=r$, then
$$
\ker\alpha^*=(\operatorname{im}\alpha)^0,
$$
whose dimension is $m-r$: extend a <basis> of $\operatorname{im}\alpha$ to one of $F^m$ and use the dual <basis>. Rank-nullity now gives $\operatorname{rank}\alpha^*=r$. This proves the <equality of row rank and column rank>.
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