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There are two matrices $A$ and $B$, both $n \times m$ for $m < n$. Both are rank $m$.

What is actually given are the matrices $A'A$ and $B'B$.

I have two questions:

  1. Solving for a root for $C = \sqrt{A'A}$ (i.e. finding $C$ such that $C'C = A'A$) and $D = \sqrt{B'B}$ where $C$ is $n \times m$ and so is $D$, are the only solutions are $C = UA$ for some $U$ such that $UU' = I$ and $U$ is $n \times n$? (similarly for $D$, $D = VB$ for some $V$ such that $VV' = I$ and $V$ is $n \times n$)?

  2. If that's the case, is it possible to find the two roots simultaneously such that $U=V$? I don't care if I don't actually know what is $U$ or what is $V$, as long as the roots I get for $C$ and $D$ are headed by the same $U$. (I don't think it is even possible to "identify" $U$ and $V$ in the general case, because they are not unique.)


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How can $C'C=A'A$ if $C$ is $m\times n$ and $A$ is $n \times m$ – chaohuang Jul 24 '12 at 14:28
The matrices $C$ and $D$ are not be rectangular, they are be square. So we cannot find matrices $C$ and $D$ of the form you prescribe. We are calculating the square roots of square matrices, as $A'A\in\mathbb{R}^{m\times m}$ then $C\in\mathbb{R}^{m\times m}$. – Andrew Winters Jul 24 '12 at 14:49
Then $C=A$, $U=I$? – chaohuang Jul 24 '12 at 15:34

The answer to your first question is negative (and hence there is no need to answer the second question). Consider $A=\pmatrix{1&1\\ 0&1\\ 0&0}$ and $C=\pmatrix{\frac{\sqrt{3}+1}2&\frac{\sqrt{3}-1}2\\ \frac{\sqrt{3}-1}2&\frac{\sqrt{3}+1}2\\ 0&0}$. Then $A^TA=C^TC=\pmatrix{2&1\\ 1&2}$. If $C$ was $UA$ for some unitary matrix $U$, both $A$ and $C$ would have identical singular values. Yet the singular values of $A$ are the golden ratio and its reciprocal, and the singular values of $C$ are $\sqrt{3}$ and $1$.

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