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'square root of positive definite matrix'
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| Title of object: |
square root of positive definite matrix |
| Canonical Name: |
SquareRootOfPositiveDefiniteMatrix |
| Type: |
Definition |
| Created on: |
2005-05-17 15:24:37 |
| Modified on: |
2006-02-06 23:50:50 |
| Classification: |
msc:15A48 |
Preamble:
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Content:
Suppose $M$ is a positive definite Hermitean 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 $M$.
We can now define the \emph{squar{e} roo{t}} of $M$ as the matrix
$$
M^{1/2}= P^T \operatorname{diag}(\sqrt{\lambda_1}, \ldots, \sqrt{\lambda_n}) P.
$$
The following properties are clear
\begin{enumerate}
\item $M^{1/2} M^{1/2}=M$,
\item $M^{1/2}$ is Hermitean and positive definite.
\item $M^{1/2}$ and $M$ commute
\item $(M^{1/2})^T=(M^T)^{1/2}$.
\item $(M^{1/2})^{-1}=(M^{-1})^{1/2}$, so one can write $M^{-1/2}$
\item If the eigenvalues of $M$ are $(\lambda_1, \ldots, \lambda_n)$, then
the eigenvalues of $M^{1/2}$ are
$(\sqrt{\lambda_1}, \ldots, \sqrt{\lambda_n})$.
\end{enumerate} |
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