I am having trouble proving the following statement.
Let $X$ and $Y$ be square two full column-rank matrices. Show that $\operatorname{rank}(XGY^T) = \operatorname{rank}(G).$
I have an idea, which is, to use SVD on $G$, which would decompose $XGY^T$ into $XUDV^TY^T$, where $U$ and $V$ are orthogonal matrices of size $m \times m$ and $n \times n$, respectively. $D$ would be an $r \times r$ matrix equivalent to $\operatorname{diag} (\sigma_1, \sigma_2, \ldots, \sigma_r)$ where $r$ is the rank of $G$. I am not sure how to proceed though, I don't understand why the idea of $X$ and $Y$ being full column-rank gets involved.