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'skew-Hermitian matrix'
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| Title of object: |
skew-Hermitian matrix |
| Canonical Name: |
SkewHermitianMatrix |
| Type: |
Definition |
| Created on: |
2003-04-29 06:32:10 |
| Modified on: |
2006-06-14 15:53:08 |
| Classification: |
msc:15A57 |
| Synonyms: |
skew-Hermitian matrix=anti-Hermitian matrix |
Preamble:
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\usepackage{amssymb}
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%\usepackage{psfrag}
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Content:
\newcommand{\ccj}[1]{\overline{#1}}
\def\dtra{\hspace{0.04cm} ^{\mbox{\scriptsize{T}}} \hspace{0.02cm}}
{\bf Definition.} A square matrix $A$ with complex entries is
\emph{skew-Hermitian}, if
$$ A^* = -A. $$
Here $A^\ast=\ccj{A\dtra}$, $A\dtra$ is the transpose of $A$, and $\ccj{A}$ is
is the complex conjugate of the matrix $A$.
\subsection*{Properties.}
\begin{enumerate}
\item The trace of a skew-Hermitian matrix is \PMlinkid{imaginary}{2017}.
\item The eigenvalues of a skew-Hermitian matrix are
\PMlinkid{imaginary}{2017}.
\end{enumerate}
\emph{Proof.} Property (1) follows directly from property (2) since the
trace is the sum of the eigenvalues. But one can also give a simple proof
as follows. Let $x_{ij}$ and $y_{ij}$ be the
real respectively imaginary parts of the elements in $A$.
Then the diagonal elements of $A$ are of the
form $x_{kk} + i y_{kk}$, and the diagonal elements in $A^\ast$
are of the form $-x_{kk} + iy_{kk}$. Hence $x_{kk}$, i.e., the real
part for the diagonal elements in $A$ must vanish, and
property (1) follows.
For property (2), suppose
$A$ is a skew-Hermitian matrix, and $x$ an
eigenvector corresponding to the eigenvalue $\lambda$, i.e.,
\begin{eqnarray}
\label{gugg}
Ax &=& \lambda x.
\end{eqnarray}
Here, $x$ is a complex column vector.
Since $x$ is an eigenvector, $x$ is not the zero vector, and
$x^\ast x > 0$. Without loss of generality we can assume $x^\ast x =1$.
Thus
\begin{eqnarray*}
\ccj{\lambda} &=& x^\ast \ccj{\lambda} x\\
&=& ( x^\ast \lambda x )^\ast \\
&=& (x^\ast A x )^\ast \\
&=& x^\ast A^\ast x \\
&=& x^\ast (-A) x \\
&=& -x^\ast \lambda x \\
&=& - \lambda .
\end{eqnarray*}
Hence the eigenvalue $\lambda$ corresponding
to $x$ is \PMlinkid{imaginary}{2017}. $\Box$ |
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