Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
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
add comment

1 Answer 1

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 $1$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.