square root of positive definite matrix

Suppose $M$ is a positive definite Hermitian matrix. Then $M$ has a diagonalization

$$M=P^{*}\mathrm{diag}({\lambda }_1,\mathrm{\dots },{\lambda }_n)P$$

where $P$ is a unitary matrix and${\lambda }_1,\mathrm{\dots },{\lambda }_n$ are the eigenvalues of $M$, which are all positive.

We can now define the square root of $M$ as the matrix

$$M^{1/2}=P^{*}\mathrm{diag}(\sqrt{{\lambda }_1},\mathrm{\dots },\sqrt{{\lambda }_n})P.$$

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 $({\lambda }_1,\mathrm{\dots },{\lambda }_n)$, thenthe eigenvalues of $M^{1/2}$ are$(\sqrt{{\lambda }_1},\mathrm{\dots },\sqrt{{\lambda }_n})$.