Near-orthonormal residual data enclose all ordered Hermitian eigenvalues

Put $G=V^{\ast }V$. For every unit vector $x$, [F6] gives $\mathrm{\mid }{x}^{\ast }(G-I)x\mathrm{\mid }\le \mathrm{\parallel }x{\mathrm{\parallel }}_2\mathrm{\parallel }(G-I)x{\mathrm{\parallel }}_2\le \mathrm{\parallel }G-I{\mathrm{\parallel }}_2\le \delta$ by [F5], so all eigenvalues $\mu$ of the Hermitian $G$ lie in $[1-\delta ,1+\delta ]$. Since $\delta <1$, $G$ is positive definite. By [F1] and [F2] it has a positive square root $H$; its eigenvalues are $\sqrt{\mu }>0$, so $H$ is invertible.

Set $Q=V{H}^{-1}$. Then $Q^{\ast }Q=H^{-1}{V}^{\ast }V{H}^{-1}=I$. Thus $Q$ is injective and [F4] makes the square matrix $Q$ invertible, so $Q$ is unitary. The eigenvalues of $H$ lie between $\sqrt{1-\delta }$ and $\sqrt{1+\delta }$, hence $\mathrm{\parallel }V-Q{\mathrm{\parallel }}_2=\mathrm{\parallel }Q(H-I){\mathrm{\parallel }}_2=\mathrm{\parallel }H-I{\mathrm{\parallel }}_2\le 1-\sqrt{1-\delta }\le \tau$. For the upper deviation, $\sqrt{1+\delta }-1\le 1-\sqrt{1-\delta }$ because $\sqrt{1+\delta }+\sqrt{1-\delta }\le 2$. Unitary invariance in [F5] justifies the norm equality.

For each $A\in \mathcal{A}$, the residual identity $AQ-Q\mathrm{\Theta }=A(Q-V)+(AV-V\mathrm{\Theta })+(V-Q)\mathrm{\Theta }$ and certificate bounds give $\mathrm{\parallel }AQ-Q\mathrm{\Theta }{\mathrm{\parallel }}_2\le M_A\tau +\rho +\tau M_{\mathrm{\Theta }}=\eta$. By the triangle inequality and submultiplicativity [F7], each product term in the identity is bounded by its factor norms, giving the displayed estimate. Since $Q$ is unitary, [F5] gives that $Q^{\ast }AQ$ has the same ordered eigenvalues as $A$ and, using [F7], $\mathrm{\parallel }{Q}^{\ast }AQ-\mathrm{\Theta }{\mathrm{\parallel }}_2=\mathrm{\parallel }{Q}^{\ast }(AQ-Q\mathrm{\Theta }){\mathrm{\parallel }}_2\le \eta$.

Both $Q^{\ast }AQ$ and $\mathrm{\Theta }$ are Hermitian. Their eigenvalues are ordered increasingly in the statement of the enclosure, whereas [F3] uses decreasing order; reversing both lists sends index $k$ to $n+1-k$, so the same absolute-difference bound holds at each increasing index. Applying [F3] to their difference gives $\mathrm{\mid }{\lambda }_k(A)-{\theta }_{(k)}\mathrm{\mid }\le \eta$ for every $k$. The family bounds are uniform, so the same intervals hold for every $A\in \mathcal{A}$.