trace of a matrixTraceOfAMatrix2001-11-16 22:30:342006-07-19 11:35:57Definition\usepackage{amssymb}
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\usepackage{xypic}\textbf{Definition} \\
Let $A=(a_{i,j})$ be a square matrix of
order $n$.
The trace of the matrix is the sum of the main diagonal:
\begin{center} $\operatorname{trace}(A)= \sum\limits _{i=1} ^{n}
a_{i,i}$
\end{center}
\textbf{Notation:}\\
The trace of a matrix $A$ is also commonly denoted as $\operatorname{Tr}(A)$
or $\operatorname{Tr}A$.
\textbf{Properties:}
\begin{enumerate}
\item The trace is a linear transformation from the space of square matrices to
the real numbers. In other words, if $A$ and $B$ are square matrices with real (or complex) entries,
of same order and $c$ is a scalar, then
\begin{eqnarray*}
\operatorname{trace}(A+B) &=& \operatorname{trace}(A)+ \operatorname{trace}(B), \\
\operatorname{trace}(cA) &=& c\cdot \operatorname{trace}(A).
\end{eqnarray*}
\item For the transpose and conjugate transpose, we have for any
square matrix $A$ with real (or complex) entries,
\begin{eqnarray*}
\operatorname{trace} (A^t) &=& \operatorname{trace} (A), \\
\operatorname{trace} (A^\ast) &=& \overline{\operatorname{trace} (A)}.
\end{eqnarray*}
\item If $A$ and $B$ are matrices such that $AB$ is a square matrix, then
$$ \operatorname{trace} (AB) = \operatorname{trace} (BA).$$
For this reason it is possible to define the trace of a linear transformation, as the choice of basis does not affect the trace.
Thus, if $A,B,C$ are matrices such that $ABC$ is a square matrix, then
$$ \operatorname{trace} (ABC) = \operatorname{trace} (CAB) = \operatorname{trace} (BCA).$$
\item If $B$ is in invertible square matrix of same order as $A$, then
$$ \operatorname{trace} (A) = \operatorname{trace} (B^{-1}A B).$$
In other words, the trace of similar matrices are equal.
\item Let $A$ be a square matrix of order $n$ with real (or complex)
entries $a_{ij}$. Then
\begin{eqnarray*}
\operatorname{trace} A^\ast A &=& \operatorname{trace} A A^\ast \\
&=& \sum_{i,j=1}^n |a_{ij}|^2.
\end{eqnarray*}
Here $^\ast$ is the complex conjugate, and $|\cdot|$ is the complex modulus.
In particular, $\operatorname{trace} A^\ast A\ge 0$ with equality if and only if $A=0$.
(See the Frobenius matrix norm.)
\item Various inequalities for $\operatorname{trace}$ are given in
\cite{yang}.
\end{enumerate}
See the proof of properties of trace of a matrix.
\begin{thebibliography}{9}
\bibitem{ehrlich}
The Trace of a Square Matrix. Paul Ehrlich, [online]
\PMlinkexternal{http://www.math.ufl.edu/~ehrlich/trace.html}{http://www.math.ufl.edu/~ehrlich/trace.html}
\bibitem{yang}
Z.P. Yang, X.X. Feng, \emph{A note on the trace inequality for
products of Hermitian matrix power},
Journal of Inequalities in Pure and Applied Mathematics,
Volume 3, Issue 5, 2002, Article 78,
\PMlinkexternal{online}{http://www.emis.de/journals/JIPAM/v3n5/082_02.html}.
\end{thebibliography}