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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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Is $U$ in the first line the same as $B$? Is $\Sigma$ the $m \times m$ diagonal matrix and you are solving for $X$? –  Ross Millikan Jul 20 '12 at 21:40
    
@Ross, terribly sorry. I fixed the question. Yes to both of your questions. –  kloop Jul 20 '12 at 21:45
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@copper, I am confused by your answer. I stared a bit at the equation and guessed the solution. My problem is that even though $U'U = I$, we don't have $UU' = I$ because $m < n$. So we can't multiply both sides by $U'$. –  kloop Jul 20 '12 at 21:48
    
@kloop: OOps, sorry, I didn't check your calculation. I will delete my comment momentarily. –  copper.hat Jul 20 '12 at 21:51
    
The title is rather extraordinary in that it's so general that it could be the title for almost any question, but it doesn't actually fit this particular question, since you're not in fact asking how to find a solution; you've already found one and want to know whether it's unique. Please try to choose titles that summarize the question. –  joriki Jul 20 '12 at 21:55
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2 Answers 2

up vote 2 down vote accepted

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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thanks. that seems correct. but why can we "characterize the solution set as the set of all matrices $X$ of the form ..." ? –  kloop Jul 20 '12 at 22:22
    
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 $VU=\Sigma^{-1}XAU=I$. –  joriki Jul 20 '12 at 22:35
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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.

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