Is a hermitian matrix invertible?

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?

• Take the zero matrix. Sep 12, 2017 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 $z$. Sep 12, 2017 at 14:33

Notably, $\det(H^*) = \det(H)^*$. Since $\det(H) = \det(H)^*$, we can conclude that $\det(H)$ is real.
Of course not. In all dimensions $\geq 2$, the matrix with all entries equal to $1$ is hermitian but not invertible (its rank is $1$).
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.