= Solution
{c}
The <power iteration> starts from $v_0\ne0$, normally scaled to unit <Euclidean norm>, and repeats
$$
w_{k+1}=Av_k,\qquad
\boxed{v_{k+1}=\frac{w_{k+1}}{\|w_{k+1}\|_2}.}
$$
Normalization prevents the rapidly growing or decaying magnitudes of $A^kv_0$ from obscuring its direction. The iteration requires $w_{k+1}\ne0$.
Because $A$ is a <diagonalizable matrix>, choose an <eigenbasis> $u_1,\ldots,u_n$, with corresponding <eigenvalues> $\lambda_1=\lambda,\lambda_2,\ldots,\lambda_n$, and expand $v_0=\sum_jc_ju_j$. First suppose the <dominant eigenvalue> is simple and $c_1\ne0$. Then
$$
A^kv_0
=c_1\lambda^k\left(u_1+\sum_{j=2}^n\frac{c_j}{c_1}
\left(\frac{\lambda_j}{\lambda}\right)^ku_j\right).
$$
For
$$
q=\max_{j\geq2}\left|\frac{\lambda_j}{\lambda}\right|<1,
$$
the sum in parentheses after $u_1$ is $O(q^k)$ in <Big O notation>. Consequently the <power iteration direction error> is $O(q^k)$: the iterates approach the line spanned by $u_1$. For a real <matrix>, a uniquely dominant <eigenvalue> is real, since otherwise its complex conjugate would have the same <modulus>. If $\lambda>0$, $v_k$ approaches one unit <eigenvector>; if $\lambda<0$, successive $v_k$ alternate signs. In either case,
$$
\boxed{\min_{\sigma\in\{-1,1\}}
\left\|v_k-\sigma\frac{u_1}{\|u_1\|_2}\right\|_2=O(q^k).}
$$
This is the <sign and phase in power iteration> qualification. Also, the necessary assumption is a nonzero coefficient $c_1$ in the <eigenbasis>, rather than merely a nonzero ordinary <inner product> with $u_1$: those tests can differ for a <nonsymmetric matrix>. If $c_1=0$, the method cannot generate the missing component.
If the dominant value $\lambda$ has a higher-dimensional <eigenspace>, the same argument replaces $c_1u_1$ by the nonzero sum of all starting components in that <eigenspace>. The limiting direction is then that of this sum, which is still an <eigenvector> for $\lambda$. The usual algorithm assumes $\lambda\ne0$; a strict dominance gap guarantees this whenever another <eigenvalue> is present.
An estimate retaining the sign of $\lambda$ is the <Rayleigh quotient>
$$
\boxed{\widehat\lambda_k=\frac{v_k^TAv_k}{v_k^Tv_k}\longrightarrow\lambda.}
$$
Continuity of this quotient at a nonzero <eigenvector> proves the limit even when $A$ is a <nonsymmetric matrix>. Alternatively, $(Av_k)_i/(v_k)_i\to\lambda$ for a coordinate whose limiting <eigenvector> component is nonzero. The scaling factor $\|Av_k\|_2$ alone tends only to $|\lambda|$.
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