4
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ $A^k$ and $D$ are similar. Similar matrices always have the same trace. $\endgroup$ Jun 6, 2013 at 12:21

2 Answers 2

7
$\begingroup$

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 $$ \tr(S^{-1}DS) = \tr(DSS^{-1}) = \tr(D). $$

$\endgroup$
2
$\begingroup$

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 $S S^{-1} = I$.

$\endgroup$
2
  • 1
    $\begingroup$ What definition of trace are you referring to? In my book the trace of a matrix is just the sum of its diagonal entries, and this definition does not involve (but implies) "being cyclic". I am aware that, like the determinant, there are sophisticated ways of characterizing the trace, but one should not confuse characterisation and definition. $\endgroup$ Jun 6, 2013 at 12:30
  • $\begingroup$ @Marc - What I mean was that if you ask for a linear operator that is cyclic, you get the trace. You've got me though, I can't find a reference for it at the moment. $\endgroup$
    – user81204
    Jun 6, 2013 at 12:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .