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. 1.

$M^{1/2}M^{1/2}=M$,

2. 2.

$M^{1/2}$ is Hermitian and positive definite.

3. 3.

$M^{1/2}$ and $M$ commute

4. 4.

$(M^{1/2})^{T}=(M^{T})^{1/2}$.

5. 5.

$(M^{1/2})^{-1}=(M^{-1})^{1/2}$, so one can write $M^{-1/2}$

6. 6.

If the eigenvalues of $M$ are $(\lambda_{1},\ldots,\lambda_{n})$, then the eigenvalues of $M^{1/2}$ are $(\sqrt{\lambda_{1}},\ldots,\sqrt{\lambda_{n}})$.

Title square root of positive definite matrix SquareRootOfPositiveDefiniteMatrix 2013-03-22 15:16:42 2013-03-22 15:16:42 rspuzio (6075) rspuzio (6075) 12 rspuzio (6075) Definition msc 15A48 CholeskyDecomposition