self-dualSelfDual2002-02-26 09:04:102002-02-26 09:05:42Definitionanti self-dual\usepackage{amsmath}
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\newtheorem{proposition}{Proposition}\paragraph{Definition.} Let $U$ be a finite-dimensional inner-product space
over a field $\kfield$. Let $T:U\rightarrow U$ be an endomorphism,
and note that the adjoint endomorphism $T\adj$ is also an endomorphism
of $U$. It is therefore possible to add, subtract, and compare $T$
and $T\adj$, and we are able to make the following definitions. An
endomorphism $T$ is said to be {\em self-dual} (a.k.a. {\em
self-adjoint}) if
$$T=T\adj.$$
By contrast, we say that the endomorphism is {\em anti self-dual} if
$$T=-T\adj.$$
Exactly the same definitions can be made for an endomorphism of
a complex vector space with a Hermitian inner product.
\paragraph{Relation to the matrix transpose.} All of these definitions have
their counterparts in the matrix setting. Let $M\in
\Mat_{n,n}(\kfield)$ be the matrix of $T$ relative to an orthogonal
basis of $U$. Then $T$ is self-dual if and only if $M$ is a symmetric matrix,
and anti self-dual if and only if $M$ is a skew-symmetric matrix.
In the case of a Hermitian inner product we must replace the transpose
with the conjugate transpose. Thus $T$ is self dual if and only if $M$ is a Hermitian matrix, i.e.
$$M = \overline{M^t}.$$
It is anti self-dual if and only if
$$M = -\overline{M^t}.$$