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Suppose we have $MM^T = NN^T$, where $M$ and $N$ are both $n$ by $d$ matrices. Assume that $n$ is (much) larger than $d$, are there anything we could conclude about $M$ and $N$, aside from that $N$ could be a permutation of $M$?

In another word, if given a $n$ by $d$ dimension matrix $M$ and $MM^T = X$, can we efficiently find a matrix $N$ such that $N$ has the same dimension as $M$ and also $NN^T = X$ (where $N$ can't be a permutation of $M$'s columns)?

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  • $\begingroup$ You say "In another word", but the two questions of what you know about solutions, and how you can generate solutions, seem to be different. $\endgroup$
    – LSpice
    Commented Apr 2 at 19:46
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    $\begingroup$ It's a standard result that the exact relationship between $M$ and $N$ is that they're related by a real unitary (sometimes called real orthogonal) matrix. That is, there's a real $d \times d$ unitary matrix $U$ such that $M = NU$. When $U$ is a permutation matrix, you get the permutation fact that you noted. When $U$ is unitary but not a permutation, you get something else (which seems to be what you want). $\endgroup$
    – Nathaniel Johnston
    Commented Apr 2 at 19:55

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