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Let $A$ and $B$ be recangular real matrices of the same shape.

Question 1. Is there any connection between $trace(AB^\top B A^\top)$ and $trace(A^\top BB^\top A)$ ?

The answer is probably, No! in general (an exception being when one of $A$ and $B$ is proportional to an orthogonal projection matrix), but maybe I'm missing something.

Question 2. What is a general condition between $A$ and $B$, for which the answer to Question 1 is affirmative ?

dohmatob
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  • Thanks for the input. There seems to be a typo. Indeed, $AB$ may not make sense since $A$ and $B$ are rectangular of the same shape. – dohmatob Jul 01 '21 at 12:50
  • Oh, you're right. Disregard my comment. – Ben Grossmann Jul 01 '21 at 12:53
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    A possibly useful observation: $$ \operatorname{tr}(AB^TBA^T) = \operatorname{tr}([AB^T][AB^T]^T) = |AB^T|_F^2,\ \operatorname{tr}(A^TBB^TA) = \operatorname{tr}([A^TB][A^TB]^T) = |A^TB|_F^2 = |B^TA|_F^2, $$ where $|\cdot|_F$ denotes the Frobenius norm. – Ben Grossmann Jul 01 '21 at 12:55
  • @BenGrossmann very interesting observation! Thanks!. May be another equivalent formulation: Is it true to say: $tr(B^TBA^TA)=tr(BB^TAA^T)$? tp have this equivalent formulation we use the cyclic property of the trace: $tr(ABC)=tr(BCA)=tr(CAB)$. So puting $ A:=A. B:=B^TB,C:=A^T$ we have $trace(AB^TBA^T)=trace(B^TBA^TA)$ – Ali Taghavi Jul 01 '21 at 13:38
  • @BenGrossmann Yes, the problem can equivalently rewritten as you've done (actually, this is the initial form of my problem; I wrote it here in terms of traces to reach a larger audience). Ali Taghavi: Yes, that is indeed correct, and that's what I used to proof my side comment that the answer to Q1 is affirmative if $A$ or $B$ is proportional to an orthogonal projection matrix. – dohmatob Jul 01 '21 at 13:44
  • Perhaps it might be helpful to consider the case where $A$ and $B$ have the same left and / or right singular vectors ? – dohmatob Jul 01 '21 at 13:46
  • @BenGrossmann One of your own questions from 2019 appears to be relevant: https://math.stackexchange.com/questions/3481667 – Hyperplane Jul 01 '21 at 16:18
  • @Hyperplane Well spotted! I forgot that I had asked that. – Ben Grossmann Jul 01 '21 at 16:28

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Following up on Ben's observation, given SVDs $A=U\Sigma V'$ and $B = R\Lambda S'$ we have

$$\begin{aligned} \|AB'\|_F^2 &= \|U\Sigma V'S\Lambda'R'\|_F^2 = \|\Sigma V'S\Lambda'\|_F^2 \\ \|B'A\|_F^2 &= \|S\Lambda'R'U\Sigma V'\|_F^2 = \|\Lambda'R'U\Sigma \|_F^2 \\ \end{aligned}$$

from this, if both $V'S$ and $R'U$ are diagonal, both norms are the same. This happens for instance when $S=V$ and $R=U$, i.e. precisely when $A$ and $B$ are simultaneously SVD-decomposable.

Recall that two square matrices are simultaneously diagonalizable if and only if they commute. It appears there is a similar condition for the existence of a simultaneous SVD decomposition:

Prove that if $A^TB$ and $AB^T$ are symmetric, then there exist orthogonal $U,V$ such that $UAV^T$ and $UBV^T$ are diagonal

However, the cited paper states that:

Either one (but in general, not both) of the diagonal matrices may be further restricted to have no negative elements

So it is not quite a simultaneous SVD, but close enough for our purposes here.

Hyperplane
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