Question: Given the singular decomposition $A=U\varSigma V^{\top}$, then $A^{\top}A=V\left( \varSigma ^{\top}\varSigma \right) V^{\top}$, $AA^{\top}=U\left( \varSigma ^{\top}\varSigma \right) U^{\top}$. Since $A^{\top}A$, $AA^{\top}$ are both symmetric, they must both have a well-defined spectral decomposition; $A^{\top}A$, $AA^{\top}$ will also have the same non-zero eigenvalues. Therefore, if $AA^{\top}=Q_1{\varLambda Q_1}^{\top}$, $A^{\top}A=Q_2\varLambda Q_{2}^{\top}$, then we can write the SVD by noting $V=Q_2$, $U=Q_1$.
Except that this forgets to determine $\sigma _i$, the obvious false is that $P, Q$ is uncertain.
However, I wonder at what condition, $P, Q$ is certain and $V=Q_2$, $U=Q_1$.
THX ♪(･ω･)ﾉ