Given regular matrices $A\_i,B\_i \in \textrm{GL}\_n(\mathbb{R}),$ $i=1,2$.


Let $A\_1 = U\_1 B\_1 V\_1$ and $A\_2=U\_2 B\_2 V\_2$ where $U\_i,V\_i \in \textrm{O}\_n(\mathbb{R})$ $(i=1,2)$ are orthogonal matrices. This means that $A\_1$ and $B\_1$ resp. $A\_2$ and $B\_2$ have the same singular values.


Now let $A\_2 A\_1^{-1}$ and $B\_2 B\_1^{-1}$ have the same singular values.


I assume that in this case one can find $U\_i',V' \in \textrm{O}\_n(\mathbb{R})$ so that $A\_1 = U\_1' B\_1 V$ and $A\_2=U\_2' B\_2 V$.


Is this true, at least under some conditions? Or is there a simple counterexample?



singular value decomposition
Let $A\_2A\_1^{-1}=U\_A DV\_A$ and $B\_2B\_1^{-1}=U\_B DV\_B$ be the two SVDs with the same $D$.


Set $U\_2'=U\_AU\_B'$, $U\_1=V\_B'V\_A$, $V=B\_1^{-1} U\_1 A\_1$.


The first equality is clear. The second one is proved by $$U\_2'B\_2V=U\_A U\_B'B\_2 B\_1^{-1} V\_B' V\_A A\_1=U\_ADV\_A A\_1=A\_2A\_1^{-1}A\_1=A\_2.$$


