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 $({\lambda}_{1},\mathrm{\dots},{\lambda}_{n})$, then the eigenvalues of ${M}^{1/2}$ are $(\sqrt{{\lambda}_{1}},\mathrm{\dots},\sqrt{{\lambda}_{n}})$.
Title  square root of positive definite matrix 

Canonical name  SquareRootOfPositiveDefiniteMatrix 
Date of creation  20130322 15:16:42 
Last modified on  20130322 15:16:42 
Owner  rspuzio (6075) 
Last modified by  rspuzio (6075) 
Numerical id  12 
Author  rspuzio (6075) 
Entry type  Definition 
Classification  msc 15A48 
Related topic  CholeskyDecomposition 