Simplifying a matrix expression that involves a Moore-Penrose pseudoinverse

Question

For $m\geq n$, suppose $L$ is an $m\times n$ real matrix and $P$ is an $m\times m$ diagonal matrix with nonnegative entries. I did some numerical experimentation to verify that $L'P (PLL'PLL'P)^\dagger PLL'P = (P^{1/2} L)^\dagger P^{1/2}$ where the dagger means the Moore-Penrose inverse. I found the differences are small enough to be rounding errors. I have a proof for the case $P=I$. Is there a formal proof of the general result (and references)?

Answer 1

I made a mistake somewhere and cannot find it yet (will take a look again later) but I think a reasoning like the following can get you the answer :

Note that $(AA^T)^\dagger = (A^T)^\dagger A^\dagger$, we also have the identities $A^T=A^\dagger A A^T$ and $A^\dagger=A^T(AA^T)^\dagger$. Let $S=P^{1/2} L$ and $Q=P^{1/2}SS^T$, observe that $S^T=L^T P^{1/2}$ and that $Q^T=SS^T P^{1/2}$, then \begin{align*} L^TP (PLL^T PLL^T P)^\dagger PLL^T P &=S^TP^{1/2} (P^{1/2}SS^T SS^T P^{1/2})^\dagger P^{1/2}SS^T P^{1/2}\\ &=S^\dagger SS^T P^{1/2} (P^{1/2}SS^T SS^T P^{1/2})^\dagger P^{1/2}SS^T P^{1/2}\\ &=S^\dagger Q^T (Q Q^T)^\dagger Q P^{1/2}\\ \end{align*}

This is really close to your identity, indeed $Q^T (Q Q^T)^\dagger Q$ is a projection matrix on the span of $Q^T$ which in turns is the span of $S$ and which in turn is the span of $(S^\dagger)^T$ hence $Q^T (Q Q^T)^\dagger Q =(S^\dagger)^T (S^\dagger (S^\dagger)^T)^\dagger S^\dagger$. plugin in we get \begin{align*} L^TP (PLL^T PLL^T P)^\dagger PLL^T P &= S^\dagger Q^T (Q Q^T)^\dagger Q P^{1/2}\\ &= S^\dagger (S^\dagger)^T (S^\dagger (S^\dagger)^T)^\dagger S^\dagger P^{1/2}\\ &=S^\dagger P^{1/2} \end{align*}