 Solution

We use the spectral theorem to construct a sufficiently accurate approximate eigenpair. Complete $x$ to an orthonormal eigenbasis $x,x_2,\dots ,x_n$ of the symmetric matrix $B$, with corresponding eigenvalues $\lambda ,{\lambda }_2,\dots ,{\lambda }_n$. Simplicity of $\lambda$ means that every ${\lambda }_j-\lambda$ below is nonzero. PutThe orthogonal decomposition of $Ex$ shows that $x^{T}v=0$ andWith $u_{\epsilon }=x+\epsilon v$ and ${\mu }_{\epsilon }=\lambda +\epsilon a$, direct expansion cancels every first-order term:Also $\Vert u_{\epsilon }{\Vert }_2^2=1+{\epsilon }^2\Vert v{\Vert }_2^2\ge 1$. Thus its eigenpair residual has Euclidean norm $O({\epsilon }^2)$ in Big O notation.

For completeness, a real symmetric matrix $M$ always has an eigenvalue within $\Vert (M-\mu I)u{\Vert }_2/\Vert u{\Vert }_2$ of any real $\mu$ and nonzero $u$. Indeed, writing $u={\sum }_j{c}_j{q}_j$ in an orthonormal eigenbasis with eigenvalues ${\theta }_j$ givesApply this symmetric eigenvalue residual bound to $M=B+\epsilon E$. There is an eigenvalue ${\lambda }_{\epsilon }$ such thatwhich provesThis establishes the first-order perturbation of a simple symmetric eigenvalue without presupposing differentiability of the perturbed eigenvector. If $n=1$, the sum defining $v$ is empty and the displayed eigenvalue formula is exact.