Property of Hermitian matrix

Question

How to prove that the sum of diagonal elements of Hermitian matrix equals to the sum of its eigenvalues?

Thanks much!

Answer 1

Hint: the trace is multiplicative, i.e. $\mbox{Tr}(AB)=\mbox{Tr}(BA)$, as soon as the scalar field (or ring) is commutative. In particular, for every invertible matrix $P$ and every matrix $A$ in $M_n(\mathbb{R})$ or $M_n(\mathbb{C})$, we have $$ \mbox{Tr}\;(PAP^{-1})=\mbox{Tr}\; A. $$

Answer 2

Any Hermitian matrix can be diagonalised, and then note that $\operatorname{tr}(P^{-1} M P) = \operatorname{tr}(M)$