Originally Posted by

**NeTTuR** I was thinking something like this:

$\displaystyle A=A^* \Rightarrow A = U D U^*$, where D diagonal matrix with eigenvalues of A and U unitary.

$\displaystyle A^2B = BA^2 \Leftrightarrow UD^2U^*B=BUD^2U^* \Leftrightarrow D^2U^*BU = U^*BUD^2$

$\displaystyle U^*BU = C \Rightarrow D^2C = CD^2$

$\displaystyle D^2C =

\begin{pmatrix}

\lambda_1^2 & 0 & \cdots & 0 \\

\vdots & \vdots & \ddots & \vdots \\

0 & 0 & \cdots & \lambda_n^2

\end{pmatrix}

\begin{pmatrix}

c_{1,1} & c_{1,2} & \cdots & c_{1,n} \\

\vdots & \vdots & \ddots & \vdots \\

c_{n,1} & c_{n,2} & \cdots & c_{n,n}

\end{pmatrix}

$$\displaystyle = \begin{pmatrix}

\lambda_1^2 c_{1,1} & \lambda_1^2 c_{1,2} & \cdots & \lambda_1^2c_{1,n} \\

\vdots & \vdots & \ddots & \vdots \\

\lambda_n^2 c_{n,1} & \lambda_n^2 c_{n,2} & \cdots & \lambda_n^2 c_{n,n}

\end{pmatrix}$

$\displaystyle CD^2 =

\begin{pmatrix}

c_{1,1} & c_{1,2} & \cdots & c_{1,n} \\

\vdots & \vdots & \ddots & \vdots \\

c_{n,1} & c_{n,2} & \cdots & c_{n,n}

\end{pmatrix}

\begin{pmatrix}

\lambda_1^2 & 0 & \cdots & 0 \\

\vdots & \vdots & \ddots & \vdots \\

0 & 0 & \cdots & \lambda_n^2

\end{pmatrix}

$$\displaystyle = \begin{pmatrix}

\lambda_1^2 c_{1,1} & \lambda_2^2 c_{1,2} & \cdots & \lambda_n^2c_{1,n} \\

\vdots & \vdots & \ddots & \vdots \\

\lambda_1^2 c_{n,1} & \lambda_2^2 c_{n,2} & \cdots & \lambda_n^2 c_{n,n}

\end{pmatrix}$ Correct as far as here.

This gives $\displaystyle \lambda_1^2 = \lambda_2^2 = \cdots = \lambda_n^2$ and $\displaystyle \lambda_i$ is an eigenvalue to A. No. If all the eigenvalues of A are equal, then A would have to be a multiple of the identity matrix, and that need not be the case. What you can deduce from equations like $\displaystyle \color{red}\lambda_i^2 c_{i,j} = \lambda_j^2 c_{i,j}$ (where $\displaystyle \color{red}i\ne j$) is that **either** $\displaystyle \color{red}\lambda_i = \lambda_j$ **or** $\displaystyle \color{red}c_{i,j} = 0$.

If I take the square root of these, then $\displaystyle DC$ will commute.

So for a Hermitian matrix, there will be a positive semidefinite matrix that commutes with $\displaystyle B$ and whose square is $\displaystyle A^2$. Now because $\displaystyle A$ and therefore also $\displaystyle A^2$ is positive semidefinite, there exist an unique positive semidefinite matrix whose square is $\displaystyle A^2$ and this matrix is of course A. Yes, it's true that a positive semidefinite matrix has a unique positive semidefinite square root.