Prove the following result for Hermitian and Skew-Hermitian matrix

Question

1. If $H$ be a Hermitian matrix, prove that $\det H$ is real number.
2. If $S$ be a skew Hermitian matrix of order $n$, prove that
   (i). if $n$ be even, then $\det S$ is real number;
   (ii). if $n$ be odd, then $\det S$ is a purely imaginary number or zero.

Attempt: 1. Let $H=P+iQ$ be a Hermitian matrix, where $P,Q$ are real matrices. Then $\bar{H}^t=H\implies P^t-iQ^t=P+iQ\implies P^t=P$ and $Q^t-Q$. How can I show that $\det H$ is real?

Answer 1

Hint:

1. The eigenvalues of hermitian matrices are real, of skew hermitian matrices are purely imaginary.
2. (Skew) Hermitian matrices are diagonalizable.
3. For $A=P^{-1}DP$, we have $\det(A)=\det(D)$.
4. What is the determinant of a diagonal matrix?

If any of these steps isn't clear to you, you need to prove it!