# Prove the following result for Hermitian and Skew-Hermitian matrix

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?

• This might be helpful. – learner Mar 24 '16 at 12:36
• $$H=H^*\implies |H|=|H^*|=|(\bar H)^T|=|\bar H|=\overline{|H|}\implies |H|-\overline{|H|}=0\implies |H|\in\Bbb R$$ – learner Mar 24 '16 at 12:42
• For the second part, $$H=-H^*\implies |H|=|-H^*|=(-1)^n|(\bar H)^T|=(-1)^n|\bar H|=(-1)^n\overline{|H|}\\ \implies |H|-(-1)^n\overline{|H|}=0$$ $$~$$ If $n$ is odd, you have $|H|+\overline{|H|}=0\implies |H|=ai~,~a\in\Bbb R$. $$~$$ If $n$ is even, you have $|H|-\overline{|H|}=0\implies |H|\in\Bbb R$ – learner Mar 24 '16 at 12:49

## 1 Answer

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!

• thanks for your answer. Is there any way to show that result without using the concept 'diagonalizable' and 'eigenvalue' – MKS Mar 24 '16 at 12:49
• @MdKutubuddinSardar: Indeed it is, as learner demonstrated in the comments. A much more elegant way than the heavy machinery of my hints. – Roland Mar 24 '16 at 13:32