= Solution
Expand $v=\sum_jc_jx_j$, where $c_j=x_j^\dagger v$. By <orthonormality>,
$$
\boxed{\beta=\sum_j|c_j|^2(\lambda_j-a)(\lambda_j-b).}
$$
Since the <spectrum> avoids $[a,b]$, each $\lambda_j$ is either below $a$ or above $b$, making every product strictly positive. This proves <positivity of a quadratic polynomial across a spectral gap>:
$$
\boxed{\beta\geq\delta\|v\|^2,\qquad\delta=\min_j(\lambda_j-a)(\lambda_j-b)>0.}
$$
In particular \b[$\beta>0$ for every nonzero $v$]. For $v=0$, $\beta=0$; the printed strict claim for an arbitrary vector must exclude this case.
Back to article page
