Trace of a diagonalized matrix

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Why do I have: $Tr(SDS^{-1})=Tr(D)$?

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2 Answers

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Note that for any matrices $A$ and $B$ we have $\def\tr{\mathop{\rm Tr}}\tr(AB) = \tr(BA)$. To see this, one can argue as follows: \begin{align*} \tr(AB) &= \sum_i (AB)_{ii}\\ &= \sum_i \sum_j A_{ij}B_{ji}\\ &=\sum_j \sum_i B_{ji}A_{ij}\\ &= \sum_j (BA)_{jj}\\ &= \tr(BA) \end{align*} Hence $$ \tr(S^{-1}DS) = \tr(DSS^{-1}) = \tr(D). $$

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It's the defining property of the trace that it is cyclic; that is

\begin{align*} Tr(AB) = Tr(BA) \end{align*}

so in your case

\begin{align*} Tr(SDS^{-1}) = Tr(DSS^{-1}) = Tr(D) \end{align*} as $S S^{-1} = I$.

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