square root of positive definite matrix

Suppose $M$ is a positive definite Hermitian matrix. Then $M$ has a diagonalization $$ M= P^* \operatorname{diag}(\lambda_1, \ldots, \lambda_n) P $$ where $P$ is a unitary matrix and $\lambda_1, \ldots, \lambda_n$ are the eigenvalues of $M$ , which are all positive.

We can now define the square root of $M$ as the matrix $$ M^{1/2}= P^* \operatorname{diag}(\sqrt{\lambda_1}, \ldots, \sqrt{\lambda_n}) P. $$ The following properties are clear

1. $M^{1/2} M^{1/2}=M$ ,
2. $M^{1/2}$ is Hermitian and positive definite.
3. $M^{1/2}$ and $M$ commute
4. $(M^{1/2})^T=(M^T)^{1/2}$ .
5. $(M^{1/2})^{-1}=(M^{-1})^{1/2}$ , so one can write $M^{-1/2}$
6. If the eigenvalues of $M$ are $(\lambda_1, \ldots, \lambda_n)$ , then the eigenvalues of $M^{1/2}$ are $(\sqrt{\lambda_1}, \ldots, \sqrt{\lambda_n})$ .