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<record version="1" id="4396">
 <title>proof of properties of trace of a matrix</title>
 <name>ProofOfPropertiesOfTraceOfAMatrix</name>
 <created>2003-06-25 21:05:14</created>
 <modified>2003-06-25 21:05:14</modified>
 <type>Proof</type>
<parent id="930">trace of a matrix</parent>
 <selfproof>0</selfproof>
 <creator id="40" name="Daume"/>
 <author id="40" name="Daume"/>
 <classification>
	<category scheme="msc" code="15A99"/>
 </classification>
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 <content>\textbf{Proof of Properties:}
\begin{enumerate}
\item Let us check linearity. For sums we have
\begin{eqnarray*}
\operatorname{trace}(A+B) &amp;=&amp; \sum\limits_{i=1}^{n} (a_{i,i} + b_{i,i})\,\,\,\,\,\,\,\,\,\,\, \mbox{(property of matrix addition)}\\
&amp;=&amp;\sum\limits _{i=1} ^{n} a_{i,i} + \sum\limits _{i=1} ^{n} b_{i,i}\,\,\,\, \mbox{(property of sums)} \\
&amp;=&amp;\operatorname{trace}(A) + \operatorname{trace}(B).
\end{eqnarray*}
Similarly,
% \item Like above the trace is a linear transformation therefore this property
% is inherited. Proving that it is a linear
% transformation.
\begin{eqnarray*}
\operatorname{trace}(cA) &amp;=&amp; \sum\limits _{i=1} ^{n} c\cdot a_{i,i}\,\,\,\,\, \mbox{(property of matrix scalar multiplication)} \\
&amp;=&amp; c\cdot \sum\limits _{i=1} ^{n} a_{i,i}\,\,\,\,\, \mbox{(property of sums)} \\
&amp;=&amp; c\cdot \operatorname{trace}(A).
\end{eqnarray*}
\item The second property follows since the transpose does not alter
the entries on the main diagonal.
\item The proof of the third property follows by exchanging the
summation order. Suppose $A$ is a $n\times m$ matrix and $B$ is a $m\times n$ matrix.
Then
\begin{eqnarray*}
\operatorname{trace} AB &amp;=&amp; \sum\limits_{i=1}^{n} \sum\limits_{j=1}^{m} A_{i,j} B_{j,i} \\
&amp;=&amp; \sum\limits_{j=1}^{m} \sum\limits_{i=1}^{n} B_{j,i} A_{i,j} \,\,\,\,\mbox{(changing summation order)}\\
&amp;=&amp; \operatorname{trace} BA.
\end{eqnarray*}
\item The last property is a consequence of Property 3 and the fact that matrix multiplication is
associative;
\begin{eqnarray*}
\operatorname{trace} (B^{-1} A B) &amp;=&amp; \operatorname{trace} \big((B^{-1} A) B\big) \\
&amp;=&amp; \operatorname{trace} \big(B(B^{-1} A) \big) \\
&amp;=&amp; \operatorname{trace} \big( (BB^{-1}) A \big) \\
&amp;=&amp; \operatorname{trace} (A).
\end{eqnarray*}

\end{enumerate}</content>
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