A Hermitian Ritz residual certifies spectral proximity and a gap-conditioned direction

Choose an orthonormal basis $f_1,\dots ,f_N$ by [F3]. For $x={\sum }_j{c}_j{f}_j$ with $\mathrm{\parallel }x\mathrm{\parallel }\le 1$, orthonormality gives $\mathrm{\mid }{c}_j\mathrm{\mid }\le 1$, so the triangle inequality yields $\mathrm{\parallel }Ax\mathrm{\parallel }\le {\sum }_j\mathrm{\parallel }A{f}_j\mathrm{\parallel }$. Thus $A$ maps the closed unit ball to a bounded set. Its closure is closed and bounded in finite-dimensional coordinates, hence compact by [F3]. The compact-operator unit-ball criterion in [F3] makes $A$ compact; Hermitian symmetry makes it self-adjoint.

Apply [F1] to the compact self-adjoint operator from step 1.1 to obtain an orthonormal eigenbasis $(e_j{)}_{j=1}^N$ with real eigenvalues ${\lambda }_j$. Expand $u={\sum }_{j=1}^N{c}_j{e}_j$, where $c_j=(u,e_j)$ and ${\sum }_j\mathrm{\mid }{c}_j{\mathrm{\mid }}^2=\mathrm{\parallel }u{\mathrm{\parallel }}^2=1$ by [F2]. Hence $\mu ={\sum }_j{\lambda }_j\mathrm{\mid }{c}_j{\mathrm{\mid }}^2\in \mathbb{R}$.

Orthogonality gives ${\rho }^2=\mathrm{\parallel }(A-\mu I)u{\mathrm{\parallel }}^2={\sum }_{j=1}^N\mathrm{\mid }{\lambda }_j-\mu {\mathrm{\mid }}^2\mathrm{\mid }{c}_j{\mathrm{\mid }}^2\ge \mathrm{dist}(\mu ,\mathrm{\Sigma }(A){)}^2{\sum }_j\mathrm{\mid }{c}_j{\mathrm{\mid }}^2$. Since the coefficient squares sum to one by step 2.1, taking nonnegative square roots proves the spectral-proximity bound.

If there is no eigenvalue other than the simple target, then $N=1$ and $u$ is a unit multiple of $e$, so the angle is zero. Otherwise choose the indexing so $e_k=e$ and ${\lambda }_k=\lambda$. The phase-invariant angle obeys ${\sin }^2\mathrm{\angle }(u,e)=1-\mathrm{\mid }{c}_k{\mathrm{\mid }}^2={\sum }_{j\mathrm{\ne }k}\mathrm{\mid }{c}_j{\mathrm{\mid }}^2$. For each $j\mathrm{\ne }k$, $\mathrm{\mid }{\lambda }_j-\mu \mathrm{\mid }\ge \mathrm{\Delta }$ by definition. Therefore ${\rho }^2={\sum }_j\mathrm{\mid }{\lambda }_j-\mu {\mathrm{\mid }}^2\mathrm{\mid }{c}_j{\mathrm{\mid }}^2\ge {\mathrm{\Delta }}^2{\sum }_{j\mathrm{\ne }k}\mathrm{\mid }{c}_j{\mathrm{\mid }}^2={\mathrm{\Delta }}^2{\sin }^2\mathrm{\angle }(u,e)$. Taking square roots proves the directional estimate; if the target is the only eigenvalue, simplicity gives the zero-angle branch already handled.

The proximity estimate minimizes over the entire spectrum and contains no ordering or target label. It only says that some eigenvalue lies within $\rho$ of $\mu$; without an additional spectral-location certificate it does not identify that eigenvalue as the lowest one.