Why is the trace of a hermitian matrix raised to an even power greater than or equal to 0?

Question

This property is used, for example, in https://arxiv.org/abs/2002.08387.

$$\mathrm{Tr}\left(A^{2k}\right)\geq 0$$

Answer 1

Let $A=U\Lambda U'$, where $U$ is unitary and $\Lambda$ is diagonal.

Then $\mathrm{tr}(A^{2k})=\mathrm{tr}((U\Lambda U')^{2k})=\mathrm{tr}(\Lambda ^{2k}U'^{2k}U^{2k})=\mathrm{tr}(\Lambda^{2k})\geq0$,

since even power of diagonal elements is nonnegative.

Answer 2

Because such a matrix is diagonalizable; more precisely, it is similar to a diagonal matrix such that all entries from the main diagonal are real numbers. And, if $\lambda\in\Bbb R$ and $n$ is even, $\lambda^n\geqslant0$.

Answer 3

being hermitian, $A=A^*$ so

$\mathrm{Tr}\left(A^{2k}\right) = \mathrm{Tr}\left(A^{k}A^{k}\right) = \mathrm{Tr}\left((A^{k})^*A^{k}\right) = \big \Vert A^k\big \Vert_F^2 \geq 0$