= Solution
For a fixed shift $s$ that is not an <eigenvalue>, <inverse iteration> is the <power method> applied to $(A-sI)^{-1}$. Before normalization, the $k$th vector is proportional to
$$
(A-sI)^{-k}x^{(0)}
=\sum_{i=1}^n c_i(\lambda_i-s)^{-k}w_i.
$$
If $s<\lambda_1$, factor out the first coefficient:
$$
(A-sI)^{-k}x^{(0)}
=c_1(\lambda_1-s)^{-k}
\left[w_1+
\sum_{i=2}^n\frac{c_i}{c_1}
\left(\frac{\lambda_1-s}{\lambda_i-s}\right)^kw_i\right].
$$
Every ratio in parentheses lies strictly between zero and one. The bracket therefore tends to $w_1$, and normalization in the <Euclidean norm> gives
$$
\boxed{x^{(k)}\longrightarrow\operatorname{sgn}(c_1)w_1.}
$$
This proves the stated convergence to $w_1$ or $-w_1$ and exhibits the <convergence of fixed-shift inverse iteration> rate.
Now suppose $\lambda_m<s<\lambda_{m+1}$. The closest eigenvalue to $s$ is one of $\lambda_m$ and $\lambda_{m+1}$.
* If $s-\lambda_m<\lambda_{m+1}-s$, the direction approaches $w_m$, but $(\lambda_m-s)^{-1}<0$, so consecutive iterates alternate signs.
* If $s-\lambda_m>\lambda_{m+1}-s$, the iterates approach one fixed sign of $w_{m+1}$ because $(\lambda_{m+1}-s)^{-1}>0$.
* At the midpoint, the two dominant inverse eigenvalues have equal modulus and opposite signs. Generic iterates alternate between the two normalized combinations proportional to $c_mw_m+c_{m+1}w_{m+1}$ and $-c_mw_m+c_{m+1}w_{m+1}$.
Finally, if $\lambda_n<s$, the closest eigenvalue is $\lambda_n$ and its inverse eigenvalue is negative. Thus the direction approaches the line spanned by $w_n$, while normalized vectors alternate between the two signs. These cases are the <sign behavior of fixed-shift inverse iteration>.
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