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I am interested in expressing the matrix $B$ that satisfies $$ (AB)^T (AB) = A^TA $$ In terms of $A$, where $A$ and $B$ are both real-valued square matrices.

Since $(AB)^T = B^T A^T$ and defining $G = A^T A$ we can rewrite this as

$$ B^T G B = G $$ But I have no idea how to continue beyond this, or if it is even possible to express $B$ in terms of $A$.

ydd
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    Well, $B$ the identity matrix certainly works. Are you looking for something more fancy? – Pedro Oct 26 '24 at 23:10
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    If $\det A\ne 0$, then you're looking exactly at the matrices $B$ such that $ABA^{-1}\in O(n)$, and I believe that this is as easy as it gets. If $\det A=0$, then I don't know much about the orthogonal group of the symmetric bilinear form with signature $(m,0,n-m)$ or how to work the pseudo-inversy stuff. – Sassatelli Giulio Oct 26 '24 at 23:14
  • @Pedro yes sorry I should've clarified. Ideally I would like a general expression for $B$ in terms of $A$ (so it describes all the possible values of $B$ that satisfy the first equation), but I have a feeling that is not possible (though I really don't know). – ydd Oct 26 '24 at 23:18
  • @SassatelliGiulio Ah okay. I tried to avoid putting any background on this problem to make it as concise as possible, but each element $a_{i,j}$ of $A$ in this case are sampled from a unit normal distribution $a_{i,j} \sim \mathcal{N}(0,1)$, and the rows are then normalized to each have Euclidean norm of 1. So $A$ is almost always invertible (I think?). I am a little worried about adding this background context to the question, as I thought it might make it more confusing. – ydd Oct 26 '24 at 23:29
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    if $A$ is invertible you may take any orthogonal matrix $O$ and then make $B = A^{-1} O A$ – Will Jagy Oct 26 '24 at 23:49
  • @Will Jagy Perfect! Thank you (Thank you to Sassatelli Giulo as well). I will try to understand why this makes sense myself. – ydd Oct 27 '24 at 00:02
  • @WillJagy I hope I am not bothering you too much, but this question is strongly related to my question (this question is actually the real question I want to answer, I came across it and found it very interesting). I hope it makes sense at least why I'm looking at this simpler question first but now I'm trying to find a good selection for $O$ s.t. $AB$ is row-normalized. I don't expect this to be an easy answer however and will continue working on this myself. – ydd Oct 27 '24 at 01:29
  • You might be able to apply vectorization on both sides and rewrite to a Kronecker product to get something useful: https://en.wikipedia.org/wiki/Vectorization_(mathematics) – Calle Dec 02 '24 at 17:33
  • @Calle thanks. This actually might also be helpful for a completely separate problem I had thrown away in my "Figure out later" folder :D – ydd Dec 03 '24 at 05:50

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