Similarity invariance of trace ($\mathrm{tr}(SA{S}^{-1})=\mathrm{tr}A$)

For an invertible matrix $S$, the cyclic property of the trace gives $\mathrm{tr}(SA{S}^{-1})=\mathrm{tr}(A{S}^{-1}S)=\mathrm{tr}A$. 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.