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

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

2. The attempt at a solution
[tex]
\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*}
[/tex]
Now, [itex]XU[/itex] and [itex](VY)^T[/itex] are orthogonal matrices, because [itex]X[/itex] and [itex]Y[/itex] 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 [itex]\Sigma_r[/itex]...
 

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