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I have an invertible matrix $A$ of size $n \times n$ and a matrix $U$ of size $n \times m$, for $m < n$. Matrix $U$ is orthonormal, meaning, the rows are orthonormal vectors. I also have an $m \times m$ diagonal matrix $\Sigma$ with positive values on the diagonal.

I want to find the solution for the equation

$X A U = \Sigma.$

Since $U$ is orthonormal, then $U^T U = I$, and therefore, for $X = \Sigma U^T A^{-1} $, we have:

$X A U = \Sigma U^T A^{-1} A U = \Sigma.$

I would like to know if this is the only solution, and if so, how to show it.

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    @joriki: Sorry, I spaced out again, I (mis-)read rectangular as square. All matrices are rectangular, of course. I will delete my comment shortly.2012-07-20

2 Answers 2

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Since $\Sigma$ and $A$ are invertible, from $\Sigma^{-1}XAU=I$, we can characterize the solution set as the set of all matrices $X$ of the form $\Sigma VA^{-1}$, where $V$ is any matrix with $VU=I$. For $m\lt n$ there are matrices other than $U'$ with this property.

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    Write $V=\Sigma^{-1}XA$. Then your equation is $VU=I$. Since $\Sigma$ and $A$ are invertible, there's a one-to-one correspondence between $V$ and $X$; being invertible, $\Sigma$ and $A$ merely "translate" between the two. So for any $V$ with $VU=I$, there's exactly one solution $X=\Sigma VA^{-1}$, and conversely, for every solution $X$, your equation requires $VU=\Sigma^{-1}XAU=I$.2012-07-20
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Take $\Sigma = \begin{bmatrix} 1 \end{bmatrix}$, $A=\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$, $U = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$. Then the equation above becomes $\begin{bmatrix} x_1 & x_2 \end{bmatrix} A U = \Sigma$ and all solutions satisfy $x_1+x_2 = 1$. So the solution is not, in general, unique.