It is known that if a positive semi-definite matrix $P\in\mathbb{C}^{n\times n}$ commutes with a matrix $A\in\mathbb{C}^{n\times n}$, then so does its square root $\sqrt{P}$.
I adapted the proof of this fact to prove if $P\in\mathbb{C}^{m\times m}, Q\in\mathbb{C}^{n\times n}$ are PSD and $A\in\mathbb{C}^{m\times n}$, then we have $PA=AQ$ implies $\sqrt{P}A=A\sqrt{Q}$. I want to make sure that my proof is correct.
Proof : Say $\lambda_1,\lambda_2,\dots,\lambda_m\in\mathbb{R}_{\geq 0}$ be the eigenvalues of $P$ and similarly let $\mu_1,\dots,\mu_n$ be the eigenvalues of $Q$. Let $f\in\mathbb{C}[t]$ be a univariate polynomial satisfying $$ f(\lambda_i)=\sqrt{\lambda_i},\quad f(\mu_j)=\sqrt{\mu_j}. $$ Then we have $f(P)=\sqrt{P}$ and $f(Q)=\sqrt{Q}$. Write $$ f(t)=c_0+c_1 t+\dots+c_k t^k. $$ Then we have $$ \sqrt{P}A = f(P)A = c_0 A + c_1 P A+\dots+c_k P^k A= A(c_0+c_1Q+\dots+c_kQ^k)=Af(Q)=A\sqrt{Q} $$ which is what I wanted to prove.
Thanks for your help.