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Homesquare root of positive definite matrix

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# 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 $(\sqrt{\lambda_{1}},\ldots,\sqrt{\lambda_{n}})$.

Related:

CholeskyDecomposition

Type of Math Object:

Definition

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Reference

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## Mathematics Subject Classification

15A48*no label found*

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