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Let $A$ be a square complex matrix and let $A=U\Sigma V^*$ be a singular value decomposition. Then $A$ can be written as

$$A=U\begin{bmatrix} \Sigma K & \Sigma L\\ 0 & 0 \end{bmatrix} U^*$$ where $V^* =\begin{bmatrix} K & L\\ M & N \end{bmatrix}U^*$. Note that $KK^*+LL^*=I$.

Question:

Prove that $$A^2=A \iff \Sigma K=I_r$$

$(\Leftarrow)$ is easy by just verification.

While doing the $(\Rightarrow)$ implication, by comparing $A^2$ and $A$ I got

$\left(\Sigma K\right)^2=\Sigma K$ and $\Sigma K \Sigma L=\Sigma L$.

How to proceed further?

David
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    Thank you..I have corrected. @M.Vinay – David Jun 10 '16 at 07:20
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    If $A$ is a square matrix, aren't $U$, $\Sigma$, and $V$ of the same order as $A$? What are the dimensions of $K$ and $L$ then? – M. Vinay Jun 12 '16 at 05:54
  • You write 'the' singular value decomposition. I was of the impression that singular value decompositions were not unique. Am I mistaken, or are you using a particular singular value decomposition? – Jonas Dahlbæk Jun 12 '16 at 16:47
  • Thank you...for the observation..you are right..I have made change.@user161825 – David Jun 13 '16 at 04:15
  • Your notation suggests to me that $\Sigma$ is a positive definite matrix whose entries consist of only the strictly positive singular values of $A$, is this correct? – Jonas Dahlbæk Jun 13 '16 at 09:21
  • yes..@user161825 – David Jun 13 '16 at 09:23

1 Answers1

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What remains to verify is that $\Sigma K x=x$ for all $x$. Pick $x$. Since $\Sigma$ is positive definite, we may write \begin{align*} x&=\Sigma\Sigma^{-1}x=\Sigma(KK^* + LL^*)\Sigma^{-1}x\\ &=\Sigma K(K^*\Sigma^{-1}x)+\Sigma L(L^*\Sigma^{-1}x). \end{align*} Since you have already verified that $(\Sigma K)^2=\Sigma K$ and $\Sigma K\Sigma L=\Sigma L$, we find \begin{align*} \Sigma K x&=(\Sigma K)^2(K^*\Sigma^{-1}x)+\Sigma K \Sigma L(L^*\Sigma^{-1}x)\\ &=\Sigma K(K^*\Sigma^{-1}x)+\Sigma L(L^*\Sigma^{-1}x)\\ &= x, \end{align*} which was what we wanted.

Jonas Dahlbæk
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