Use SVD to show rank(XGY) = rank (G)

[endeavor]

1. Use the Singular Value Decomposition (SVD) of G to prove:
$$rank(XGY^T) = rank (G)$$
Given that $X$ and $Y$ are two full column-rank matrices, but may not have the same rank.

2. The attempt at a solution
$$\begin{eqnarray*} XGY^T & = & X(U\Sigma V^T)Y^T \\ & = & XU \left( \begin{array}{cc} \Sigma_{r} & 0 \\ 0 & 0 \\ \end{array} \right) V^{T}Y^T \end{eqnarray*}$$
Now, $XU$ and $(VY)^T$ are orthogonal matrices, because $X$ and $Y$ are orthogonal since they have full column rank (right?). Then somehow I want to argue that the rank of this matrix must the dimension of $\Sigma_r$...

 

[kurt.physics]

Since U and V are unitary, UΣV^T must also be unitary. Therefore, the rank of XGY^T must be equal to the rank of G, which is the dimension of Σ_r.