Problem:
Let $A$ be a Hermitian positive semi-definite $n$ by $n$ matrix (The field of scalars is $\mathbb{C}$). Let $B$ be an $n$ by $n$ matrix that commutes with A. Prove that $B$ and $\sqrt{A}$ commute.
I started to solve the problem like this: Since $A$ is a Hermitian positive semi-definite matrix, then there exits a unitary matrix $J$ such that: $J^{-1}AJ=D=diag(\lambda _{1},\lambda _{2},...,\lambda _{n})$ where: $\lambda _{i}\geq 0$ and $\lambda _{i}$ are eigenvalues of $A$, and since $A$ is Hermitian, then the eigenvalues are real.
Let $F=\sqrt{A}=JCJ^{-1}$ where $C=\sqrt{D}=diag\left ( \sqrt{\lambda _{1}},\sqrt{\lambda _{2}},...,\sqrt{\lambda _{n}} \right )$, and $\sqrt{\lambda _{i}}$ are eigenvalues of $F=\sqrt{A}$.
$AB=BA\Rightarrow JDJ^{-1}B=BJDJ^{-1}$
So, I need to use the above equality to prove this one: $FB=BF\Rightarrow JCJ^{-1}B=BJCJ^{-1}\Rightarrow ...$
Please let me know how I should finish my proof. Also, if anyone has a different solution, please share. Thanks