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Let $A$ and $B$ be two $n \times n$ full-rank matrices.

Let $XAY = B$ be the given equation where $X$ and $Y$ are unknown $n \times n$ matrices. We know that $Vec(B) = (Y^{T} \otimes X)Vec(A)$. Under what conditions can we determine $X$ and $Y$ and what would be the procedure to determine $X$ and $Y$?

Not looking for the obvious solutions such as $Y^{-1}=A$ and $X=B$, $A$ is the eigenvalue (or some multiple of eigenvalue) diagonal matrix of $B$ where $X$ and $Y$ diagonalizes $B$ and $A$ and $B$ are similar to $X$ and $Y$. In general are there any other possible choices that $Y^{T} \otimes X$ can take? These may be the only choices for general $A$ and $B$. However, given $A$ and $B$ are some other structured matrices, has such an equation been studied before?

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    $\begingroup$ $X=I$ and $Y=A^{-1}B$? or am I missing something? $\endgroup$
    – Suvrit
    Commented Aug 31, 2011 at 0:18

1 Answer 1

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Since $A$ and $B$ are invertible, replace $X$ by $X' = XA$ and replace $Y$ by $Y' = YB^{-1}$. Then your equation reduces to solving $X'Y' = I$, i.e. the space of solutions is parametrized by $GL(n)$.

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  • $\begingroup$ I was actually curious of a meaning behind Theorem 9.3.5 here: books.google.com/… $\endgroup$
    – user16007
    Commented Aug 31, 2011 at 4:23
  • $\begingroup$ No idea about that, sorry. I can't see the second half of the theorem since Google is not showing me page 311. $\endgroup$
    – MTS
    Commented Aug 31, 2011 at 4:48
  • $\begingroup$ I think you need only pages 309 and 310! $\endgroup$
    – user16007
    Commented Aug 31, 2011 at 11:27

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