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A Hermitian matrix is defined as $H=H^\dagger$. Taking the determinant on both sides, and using $\det(A)=\det(A^T)$ we get, $\det(H)=\det(H^*)$. What can we say about the determinant of $H$ from this?

SRS
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    Take the zero matrix. – mechanodroid Sep 12 '17 at 14:31
  • Note that among mathematicians, the more common convention is to use $H^$ (as opposed to $H^\dagger$) to denote the adjoint of $H$ and to use $\overline{z}$ (as opposed to $z^$) to denote the complex conjugate of $z$. – Ben Grossmann Sep 12 '17 at 14:33

3 Answers3

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Notably, $\det(H^*) = \det(H)^*$. Since $\det(H) = \det(H)^*$, we can conclude that $\det(H)$ is real.

Of course, Hermitian matrices are not generally invertible. Note, for example, that the zero-matrix is Hermitian but is certainly not invertible.

Ben Grossmann
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Of course not. In all dimensions $\geq 2$, the matrix with all entries equal to $1$ is hermitian but not invertible (its rank is $1$).

Francesco Polizzi
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The eigenvalues are real so the determinant will be real, but not much else can be said. As to invertibility $$ \left(\begin{array}{ccc} \lambda_1&0&0\\ 0&0&0 \\ 0&0&\lambda_2\end{array}\right)\, ,\qquad \lambda_1\,\lambda_2\in\mathbb{R} $$ and an infinite number of variations on this theme provide examples of non-invertible hermitian matrices.