A vanishing eigengap allows a fixed eigenvector rotation

The eigenvalues of $A_{\delta }$ are $1+\delta$ and $1$, so its leading line is $\mathrm{span}(e_1)$ and its gap is $g_{\delta }=\delta$. Also $E_{\delta }^{\mathsf{T}}{E}_{\delta }=(\delta \mathrm{/}4{)}^2{I}_2$, whence [F1] gives $\mathrm{\parallel }{E}_{\delta }{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}=\delta \mathrm{/}4$.

The characteristic polynomial of $A_{\delta }+E_{\delta }=\left(\begin{array}{cc}1+\delta & \delta \mathrm{/}4\\ \delta \mathrm{/}4& 1\end{array}\right)$ has roots $1+\delta \mathrm{/}2\pm \sqrt{(\delta \mathrm{/}2{)}^2+(\delta \mathrm{/}4{)}^2}=1+\delta \mathrm{/}2\pm \sqrt{5}\delta \mathrm{/}4$, so the larger root is the leading eigenvalue. If $(x,y{)}^{\mathsf{T}}$ is its leading eigenvector, the first row of the eigenvalue equation gives $y\mathrm{/}x=4({\lambda }_1(A_{\delta }+E_{\delta })-1-\delta )\mathrm{/}\delta =\sqrt{5}-2$. Thus its line makes angle $\arctan (\sqrt{5}-2)>0$ with $\mathrm{span}(e_1)$, independently of $\delta$.

Letting $\delta \downarrow 0$ makes $\mathrm{\parallel }{E}_{\delta }{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\to 0$ by step 1.1 while the angle in step 1.2 stays fixed and positive, contradicting the proposed uniform claim. Here ${ϵ}_{\delta }=g_{\delta }\mathrm{/}4<g_{\delta }\mathrm{/}2$, so [F2] still applies; its gap-dependent bound is consistent with the example.