When a matrix is Hermitian. Prove that when the matrix is powered to an integer, an even integer is the only case that produces a Hermitian and Positive semidefinite matrix.
if $M = M^{H}$ then $M^{e}=(M^{e})^{H}$ and $v^{T}M^{e}v \geq0$?
My Attempt:
To prove it's hermitian: (M$\times$M...M)$^{H}=(M^{H}$$\times$$M^{H}...M^{H})=(M^{H})^{n}$
If $M$ is Hermitian, then it is unitarily diagonalizable.
$$M=UDU^H$$
and $D$ consists of real eigenvalues along the diagonal and $0$ elsewhere and $U$ is unitary.
If $e$ is an even number, then
$$M^e=UD^eU^H$$
where $D^e$ will consists of nonnegative diagonal entries.
However, for an odd number, $p$, $D^p$ might consists of some negative numbers.
However, one should be cautious. Even if $p$ is odd, it is possible that for some $M$, $M^e$ is positive semidefinite, take the zero matrix or the identity matrix for example.