Notice that the row-rank is not affected by multiplication on the left or on the right by an invertible matrix (prove this, it's not very hard).

But using both row and column operations we can reduce $\displaystyle A$ to a matrix

$\displaystyle D=\begin{pmatrix}

{I_r}&{O}\\

{O}&{O}

\end{pmatrix}$

where the O's are all of appropriate size. This is equivalent to writing $\displaystyle A=PDQ$. But $\displaystyle A^T = Q^TD^TP^T = Q^TDP^T$; but then $\displaystyle \mbox{row-rank }A^T = \mbox{row-rank }Q^TDP^T = \mbox{row-rank }D = \mbox{row-rank }A$.