If two square matrices $M_1$ and $M_2$ commute, does it mean that the $U$ and $V^\dagger$ appearing in their singular value decompositions are the same?
Specifically, does it imply that
$$M_1 = U \Sigma_1 V^\dagger$$ $$M_2 = U \Sigma_2 V^\dagger$$
for two possibly different $\Sigma_{1,2}$?
As $M_1$ and $M_2$ commute, I know that they can be simultaneously diagonalized, which means there's an orthogonal transformation $\mathcal{O}$ which can diagonalize both matrices. But does this have any bearing on $U$ and $V$ of the singular value decomposition?
Counter-example: take $M_1 = I$, the identity matrix, and $M_2$ as any matrix of the same dimension (not a scalar multiple of the identity). You can see the above statement fails to hold.