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?


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 $$U_2'B_2V=U_A U_B'B_2 B_1^{-1} V_B' V_A A_1=U_ADV_A A_1=A_2A_1^{-1}A_1=A_2.$$

  • $\begingroup$ Thank you. This is exactly the kind of result I was looking for. $\endgroup$ – Döni Jul 7 '12 at 17:13

This is really a commentary, but too loong ...

I dont really understand. ¿Is the $B_i$ supposed to be diagonal, as it looks like, since the title is the "singuar value decomposition"? But then $B_i$ has the singular values of $A_i$. Assuming this, both $A_1^T A_1$ and $A_2^T A_2$ are symmetric positive definite matrices, so by simultaneous diagonalization there exists a diagonal matrix $\Lambda$ with positive entries and a non-singular matrix (not orthogonal!) so that $X^T A_1^T A_1 X=\Lambda$ and $X^T A_2^T A_2 X=I$. It follows that $(A_2 X)^T(A_2 X)=I$ so $A_2 X$ is orthogonal, write $A_2 X=V$. We have $A_1^T A_1 = X^{-T} \Lambda X^{-1} = (\Lambda^{1/2} X^{-1})^T \Lambda^{1/2} X^{-1}$ so we can conclude that $\Lambda^{1/2}X^{-1}$ is one square root of $A_1^T A_1$. $A_1$ is another square root, and since all square roots are orthogonally related, there exist some orthogonal matrix $U$ such that $A_1 = U \Lambda^{1/2} X^{-1}$ and we have already found there is some orthogonal matrix $V$ such that $A_2 = V X^{-1}$. This is close to what you have given, but the common matrix in the two expressions is not orthogonal, but general non-singular.

  • $\begingroup$ Hello Kjetil, thanks for your comments. Actually the $B_i$ are not supposed to be diagonal. The condition says: $A_i$ and $B_i$ have the same singular values. Perhaps I should rewrite the question. But of course if my assumption is true, it should also hold for diagonal matrices $B_1,B_2$. In your comment, you didn't use the fact, that $A_2 A_1^{-1}$ and $B_2 B_1^{-1}$ have the same singular values. It is clear, that if we drop this condition, the assumption will be false. $\endgroup$ – Döni Jul 7 '12 at 6:45
  • $\begingroup$ @Kjetil: I think you have rediscovered the generalized singular value decomposition (don't look at the Wikipedia page, it lists an intermediate form as the "real" result, but rather here books.google.de/…). $\endgroup$ – Federico Poloni Jul 7 '12 at 9:51
  • $\begingroup$ Yes, I think you should restate your question, the number of answers so far indicates that it is not to clear ... Be careful to include your motivation for the question, why you think this should be true ... $\endgroup$ – kjetil b halvorsen Jul 8 '12 at 19:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.