Show: If $\mathbf{A}\in\mathbb{R}^{M\times N}$ with $M\geq N$, then there exists a matrix $\mathbf{G}$ with orthonormal rows so that $\mathbf{A}^T=\mathbf{G}\mathbf{A}\mathbf{G}$.
I'm pretty lost on this problem. I know that I'm going to have to use the SVD at some point in the proof though. Does anyone have any ideas?
Assume rank of matrix $A$ to be $N$.
We can start with SVD, which provides the following decomposition: $$ A_{M\times N} = U_{M\times N} \Sigma_{N\times N} V^\top_{N\times N}. $$ Then, $$ A^\top = V\Sigma U^\top = VU^\top U \Sigma V^\top V U^\top = VU^\top A VU^\top. $$ Pattern matching with $A^\top = GAG$, gives $G = VU^\top$.
Rows of $G$ are orthonormal is equivalent to $G G^\top = I_{N\times N}$. $$ GG^\top = VU^\top UV^\top = VV^\top = I_{N\times N}. $$ The last equality is because $V$ is a square matrix and $V^\top V = I$ and therefore $V^{-1} = V^\top$.