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

Question

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?

Answer 1

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.