skew-symmetric matrixSkewSymmetricMatrix2001-11-21 21:24:502006-07-05 10:49:40Definition\usepackage{amssymb}
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\usepackage{xypic}\textbf{Definition:} \newline Let $A$ be an square matrix of
order $n$ with real entries $(a_{ij})$.
The matrix $A$ is skew-symmetric if $a_{ij} =
-a_{ji}$ for all $1 \leq i \leq n, 1 \leq j \leq n$.
\begin{center}$A =
\begin{pmatrix}
a_{11}=0 & \cdots & a_{1n} \\
\vdots & \ddots & \vdots \\
a_{n1} & \cdots & a_{nn}=0
\end{pmatrix}$
\end{center}
The main diagonal entries are zero because
$a_{i,i} = -a_{i,i}$ implies $a_{i,i} = 0$.
One can see skew-symmetric matrices as a
special case of complex skew-Hermitian matrices. Thus,
all properties of skew-Hermitian matrices also hold
for skew-symmetric matrices.
\textbf{Properties:}
\begin{enumerate}
\item The matrix $A$ is skew-symmetric if and only if
$A^t = -A$, where $A^t$ is the matrix transpose
\item For the trace operator, we have that
$\operatorname{tr}(A) = \operatorname{tr}(A^t)$.
Combining this with property (1), it follows
that $ \operatorname{tr}(A)=0$ for a skew-symmetric matrix $A$.
\item Skew-symmetric matrices form a vector space: If $A$ and $B$
are skew-symmetric and $\alpha, \beta\in \mathbb{R}$, then
$\alpha A + \beta B$ is also skew-symmetric.
\item Suppose $A$ is a skew-symmetric matrix and $B$ is a matrix of
same order as $A$. Then $B^t A B$ is skew-symmetric.
\item All eigenvalues of skew-symmetric matrices are
purely imaginary or zero. This result is proven on the page
for skew-Hermitian matrices.
\item According to Jacobi's Theorem, the determinant of a
skew-symmetric matrix of odd order is zero.
\end{enumerate}
\textbf{Examples:}
\begin{itemize}
\item $\begin{pmatrix}
0 & b \\
-b & 0
\end{pmatrix}$
\item $\begin{pmatrix}
0 & b & c \\
-b & 0 & e \\
-c & -e & 0
\end{pmatrix}$
\end{itemize}