(i) Let $A$ be an $n \times n$ symmetric real matrix with distinct eigenvalues $\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n}$ and corresponding eigenvectors $\mathbf{v}_{1}, \mathbf{v}_{2}, \ldots, \mathbf{v}_{n}$, where $\left\|\mathbf{v}_{l}\right\|=1$. Given $\mathbf{x}^{(0)} \in \mathbb{R}^{n},\left\|\mathbf{x}^{(0)}\right\|=1$, the sequence $\mathbf{x}^{(k)}$ is generated in the following manner. We set

$$\begin{gathered} \mu=\mathbf{x}^{(k) T} A \mathbf{x}^{(k)} \\ \mathbf{y}=(A-\mu I)^{-1} \mathbf{x}^{(k)} \\ \mathbf{x}^{(k+1)}=\frac{\mathbf{y}}{\|\mathbf{y}\|} \end{gathered}$$

Show that if

$$\mathbf{x}^{(k)}=c^{-1}\left(\mathbf{v}_{1}+\alpha \sum_{l=2}^{n} d_{l} \mathbf{v}_{l}\right)$$

where $\alpha$ is a real scalar and $c$ is chosen so that $\left\|\mathbf{x}^{(k)}\right\|=1$, then

$$\mu=c^{-2}\left(\lambda_{1}+\alpha^{2} \sum_{j=2}^{n} \lambda_{j} d_{j}^{2}\right)$$

Give an explicit expression for $c$.

(ii) Use the above result to prove that, if $|\alpha|$ is small,

$$\mathbf{x}^{(k+1)}=\tilde{c}^{-1}\left(\mathbf{v}_{1}+\alpha^{3} \sum_{l=2}^{n} \tilde{d}_{l} \mathbf{v}_{l}\right)+O\left(\alpha^{4}\right)$$

and obtain the numbers $\tilde{c}$ and $\tilde{d}_{2}, \ldots, \tilde{d}_{n}$.