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  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?

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  • $\begingroup$ This might be helpful. $\endgroup$ – learner Mar 24 '16 at 12:36
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    $\begingroup$ $$H=H^*\implies |H|=|H^*|=|(\bar H)^T|=|\bar H|=\overline{|H|}\implies |H|-\overline{|H|}=0\implies |H|\in\Bbb R$$ $\endgroup$ – learner Mar 24 '16 at 12:42
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    $\begingroup$ 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 $|H|-\overline{|H|}=0\implies |H|\in\Bbb R$ $\endgroup$ – learner Mar 24 '16 at 12:49
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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!

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

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