 Solution

Complete the square in the Hermitian matrix polynomial:For every nonzero vector $v$,Thus $H$ is a positive-definite matrix, so $\boxed{0\text{ is not an eigenvalue of }H}$. In finite-dimensional spectral theorem terms, each eigenvalue isIf ${\alpha }^2\ge 4\beta$, the scalar polynomial has real roots of a polynomialAn allowed nonzero root can be an eigenvalue of $A$, a Hermitian matrix which is also an invertible matrix, in which case $H$ has a zero eigenvalue. For example, $\alpha =2$, $\beta =1$, $A=-I$ gives $H=0$. The weaker inequality does not force failure for every $A$; it removes the general positive lower bound. There is also the harmless corner $\alpha =\beta =0$, for which $H=A^2$ stays positive definite because $A$ is invertible, even though the strict discriminant hypothesis is absent.