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Is there any way to solve for $X$ the equation:

$$XAX^T=B$$

Where $A$ is an $n\times n$ matrix, $B$ is an $m\times m$ matrix, and $X$ is an $m\times n$ matrix, with $m < n$. The elements of $X$ can be real or complex, I'm looking for any solution at all.

So are there any methods, research or ideas for finding a solution or a set of solutions for this problem?

b00n heT
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Chickenmales
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  • I have no clue on how to attach the problem itself, but clearly if the rank of $B$ is greater than $n$ the equation has no solution. – b00n heT Aug 26 '16 at 09:07
  • @ Chickenmales , I don't see any work; at least, consider the case $m=1,n=2$. –  Aug 28 '16 at 21:10
  • I had a look at the case where $m = 1, n = 2$ and I got $a_{ij}x_{i}x_{j} = b$ which is one equation with two variable, so there'll be one degree of freedom in the answer. – Chickenmales Aug 31 '16 at 07:02
  • In general the problem can be written as $a_{ij}x_{ai}x_{bj} = b_{ab}$ which is $mn$ variables and $m^2$ equations. so there'll be $m(n - m)$ degrees of freedom. A solution can be found by using the quadratic formula and substitution. – Chickenmales Aug 31 '16 at 07:05

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Necessarily $rank(A)\geq rank(B)$. Moreover $X(A+A^T)X^T=B+B^T$; then necessarily, $rank(A+A^T)\geq rank(B+B^T)$.

Assume that $A,B$ are real and $B+B^T$ is invertible; let $signature(B+B^T)=\{p\times +,(m-p)\times -\}$. Thus necessarily, $X^T$ is injective and $signature(A+A^T)$ is in the form $\{u\times +,v\times -,n-u-v\times 0\}$ with $u\geq p,v\geq m-p$. I think that the previous conditions suffice to the existence of $X$ s.t. $(*)$ $X(A+A^T)X^T=B+B^T$. Of course, a solution of $(*)$ is not necessarily solution of $XAX^T=B$.

EDIT . For the general solution of $(*)$ when $A+A^T\geq 0,B+B^T\geq 0$, see my answer in Is there a way to directly solve this matrix equation: $XAX^{T} = B$

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