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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 $X$?2012-07-20
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    @Ross, terribly sorry. I fixed the question. Yes to both of your questions.2012-07-20
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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 $U'$.2012-07-20
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    @kloop: OOps, sorry, I didn't check your calculation. I will delete my comment momentarily.2012-07-20
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    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.2012-07-20
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    @joriki, absolutely right. I changed the title to fit better the question.2012-07-20
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    Now the title fits the question, but it still doesn't summarize it :-) Imagine what the main page would look like if all titles were as general as this one. How about something like "uniqueness of the solution of an equation for a rectangular matrix"?2012-07-20
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    If rows of $U$ are orthonormal, we should have $UU^T=I$, right? not $U^TU=I$2012-07-20
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    @chaohuang: Yes; also the rows can't be orthonormal, since $m\lt n$, so they're linearly dependent. I think everything would be OK if "rows" is replaced by "columns" in the question.2012-07-20
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    @copper.hat: How do you mean? $X$ is $m\times n$ with $m\lt n$. In fact it's clearly rectangular in your own answer?2012-07-20
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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

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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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    thanks. that seems correct. but why can we "characterize the solution set as the set of all matrices $X$ of the form ..." ?2012-07-20
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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.