= Solution
Put
$$
M=\sum_{i\geq2}a_i^2,
\qquad
c^2=1+\epsilon^2M.
$$
The <Rayleigh quotient> shift is exactly
$$
s_k
=\frac{\lambda_1+\epsilon^2\sum_{i\geq2}a_i^2\lambda_i}
{1+\epsilon^2M}.
$$
Expanding the denominator gives
$$
s_k
=\lambda_1+\epsilon^2
\sum_{i\geq2}a_i^2(\lambda_i-\lambda_1)
+O(\epsilon^4).
$$
The question prints this as $s_k=\lambda_1-K\epsilon^2+O(\epsilon^4)$, so its sign convention requires
$$
\boxed{
K=\sum_{i\geq2}a_i^2(\lambda_1-\lambda_i)<0
}
$$
unless every $a_i$ vanishes. Equivalently, $-K$ is the positive quadratic Rayleigh-quotient error.
The shifted inverse solve has the <spectral decomposition>
$$
y=c^{-1}\left[
\frac1{\lambda_1-s_k}w_1
+\epsilon\sum_{i\geq2}
\frac{a_i}{\lambda_i-s_k}w_i
\right].
$$
Only relative coefficients matter after normalization, so factor out $(\lambda_1-s_k)^{-1}$:
$$
y\mathrel\parallel
w_1+\epsilon\sum_{i\geq2}
a_i\frac{\lambda_1-s_k}{\lambda_i-s_k}w_i.
$$
From the shift expansion,
$$
\lambda_1-s_k=K\epsilon^2+O(\epsilon^4),
$$
whereas, for $i\geq2$,
$$
\lambda_i-s_k
=(\lambda_i-\lambda_1)+K\epsilon^2+O(\epsilon^4).
$$
Consequently
$$
\epsilon a_i\frac{\lambda_1-s_k}{\lambda_i-s_k}
=\epsilon^3\frac{Ka_i}{\lambda_i-\lambda_1}
+O(\epsilon^5).
$$
After the final normalization,
$$
\boxed{
x^{(k+1)}=c_1^{-1}\left(
w_1+\epsilon^3\sum_{i\geq2}b_iw_i+O(\epsilon^5)
\right),
\qquad
b_i=\frac{Ka_i}{\lambda_i-\lambda_1}.
}
$$
\b[Thus an $O(\epsilon)$ eigenvector-direction error becomes $O(\epsilon^3)$ in one step, proving the <Local cubic convergence of Rayleigh quotient iteration> for this symmetric matrix.]
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