Similarity invariance of trace
= Similarity invariance of trace
{title2=$\operatorname{tr}(SAS^{-1})=\operatorname{tr}A$}
For an invertible <matrix> $S$, the <cyclic property of the trace> gives $\operatorname{tr}(SAS^{-1})=\operatorname{tr}(AS^{-1}S)=\operatorname{tr}A$. Thus the diagonal sum defining the <trace> does not depend on the choice of <basis> for an <endomorphism>. Cyclic permutation of factors is valid inside a <trace>; arbitrary permutation of factors need not be.