There are two matrices $A$ and $B$, both $n \times m$ for $m < n$. Both are rank $m$.
What is actually given are the matrices $A'A$ and $B'B$.
I have two questions:
Solving for a root for $C = \sqrt{A'A}$ (i.e. finding $C$ such that $C'C = A'A$) and $D = \sqrt{B'B}$ where $C$ is $n \times m$ and so is $D$, are the only solutions are $C = UA$ for some $U$ such that $UU' = I$ and $U$ is $n \times n$? (similarly for $D$, $D = VB$ for some $V$ such that $VV' = I$ and $V$ is $n \times n$)?
If that's the case, is it possible to find the two roots simultaneously such that $U=V$? I don't care if I don't actually know what is $U$ or what is $V$, as long as the roots I get for $C$ and $D$ are headed by the same $U$. (I don't think it is even possible to "identify" $U$ and $V$ in the general case, because they are not unique.)
Thanks!