Why is the trace of this object non-negative? $\phi$ is an arbitrary superposition of powers of a single hermitian matrix, $M^k$:
$$\phi =\sum_{k} \alpha_{k} M^{k}$$
Why is the following statement true?
$$\langle \mathrm{tr}\left(\phi^\dagger \phi \right)\rangle \geq 0$$
I have recently asked a related question, Why is the trace of a hermitian matrix raised to an even power greater than or equal to 0?. I understand that the trace over any $M^{2k}$ would be non-negative. But here the product $\phi^\dagger \phi$ will in general have odd powers, and so there could be negative terms in the trace, according to my understanding. 
I am coming across this statement in Eq. (3.9) of https://arxiv.org/abs/2002.08387 

Actually if $M$ is any matrix we have $\text{tr(M*M)} \geq 0$. This is clear if $M$ is diagonalizable and diagonalizable matrices are dense in all matrices.
