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Take given matrices $A, B \in \mathbb{R}^{m \times n}$. We want to find an orthogonal matrix $Q$ (in $\mathbb{R}^{n \times n}$) to maximize the inner product $\langle A, QB \rangle$.

So I know that if this problem were instead to find $Q, R$ to maximize $\langle A, QBR \rangle$, we would just take $Q = U_A U_B^T$ and $R = V_B V_A^T$ from the SVDs of $A, B$. That actually achieves the maximum $\textrm{Tr}(\Sigma_A \Sigma_B)$.

But I'm not sure how to proceed when $B$ is only multiplied from the left. I suspect that we would still take $Q = U_A U_B^T$ so the inner product is $\textrm{Tr}(\Sigma_A \Sigma_B V_B^T V_A)$, but I have no idea how to show that this is indeed the maximum. My thought is that $V_B^T V_A$ will always be in the inner product, but why?

Would really appreciate answers centered around the SVD, as well as general intutition here. Thank you!

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    $\begingroup$ This is basically an orthogonal Procrustes problem and its answer is well-known: the optimal $Q$ is $VU^T$, where $USV^T$ is a SVD of $BA^T$. $\endgroup$
    – user1551
    Commented Apr 27, 2023 at 2:26
  • $\begingroup$ Oh, that makes a ton of sense! Just gotta rotate in the trace. $\endgroup$
    – adamc
    Commented Apr 27, 2023 at 2:29
  • $\begingroup$ There appears to be a typo in the sizes of your matrices. If $\ A\ $ and $\ B\ $ are $\ m\times n\ $ matrices won't you need $\ Q\ $ to be an $\ m\times m\ $ matrix (rather than $\ n\times n\ $) for the product $\ QB\ $ to be well-defined? $\endgroup$ Commented Apr 27, 2023 at 2:40

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as mentioned above this is similar to the Procrustes problem, however, it is possible to provide another solution not based on the SVD decomposition, the complete proof can be found in this paper, in Lemma 4.2.2, page 82. I hope you find it useful.

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  • $\begingroup$ Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. $\endgroup$
    – CrSb0001
    Commented Oct 9, 2023 at 16:08

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