= Solution
Let the computed <eigenpair residual> be
$$
r=A\widetilde v-\widetilde\lambda\widetilde v,
\qquad \|r\|_2=\epsilon,
\qquad \|\widetilde v\|_2=1.
$$
Define the <rank-one matrix>
$$
E=-r\widetilde v^T.
$$
Then
$$
(A+E)\widetilde v
=A\widetilde v-r(\widetilde v^T\widetilde v)
=\widetilde\lambda\widetilde v,
$$
so $(\widetilde\lambda,\widetilde v)$ is an exact eigenpair of the perturbed matrix $A+E$. Moreover,
$$
\|E\|_2=\|r\|_2\|\widetilde v\|_2=\epsilon,
$$
which is the <backward error of an approximate eigenpair>.
Follow the simple eigenvalue branch from $\lambda(0)=\lambda$ for $A(t)=A+tE$. The <first-order perturbation of a simple eigenvalue> and part (b) give
$$
|\widetilde\lambda-\lambda|
=|\lambda(1)-\lambda(0)|
\approx|\lambda'(0)|
\leq\|E\|_2s(\lambda).
$$
Since the residual is at the <machine precision> scale, this proves
$$
\boxed{|\widetilde\lambda-\lambda|\lesssim\epsilon s(\lambda)}.
$$
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