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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.)

Thanks!

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    How can $C'C=A'A$ if $C$ is $m\times n$ and $A$ is $n \times m$2012-07-24
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    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 $C\in\mathbb{R}^{m\times m}$.2012-07-24
  • 0
    Then $C=A$, $U=I$?2012-07-24

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