= Power iteration convergence for a diagonalizable matrix
{title2=$|\lambda_1|>|\lambda_2|,\quad (V^{-1}x_0)_1\ne0$}
Suppose $A\in\mathbb C^{n\times n}$ is a <diagonalizable matrix>, $n\ge2$, with an <eigenbasis> $v_1,\ldots,v_n$ and <spectral decomposition>
$$
A=V\Lambda V^{-1},\qquad V=(v_1\ \cdots\ v_n),\qquad \Lambda=\operatorname{diag}(\lambda_1,\ldots,\lambda_n).
$$
Assume the <dominant eigenvalue> $\lambda_1$ is a <simple eigenvalue> and
$$
|\lambda_1|>|\lambda_2|\ge\cdots\ge|\lambda_n|,\qquad x_0=\sum_{j=1}^n c_jv_j,\qquad c_1\ne0.
$$
Then the normalized <power iteration> approaches the one-dimensional <eigenspace> of $\lambda_1$. Its asymptotic direction-error bound has factor $\gamma=|\lambda_2/\lambda_1|<1$. This <spectral gap> is a gap in <modulus>: if $g=|\lambda_1|-|\lambda_2|$, then $\gamma=1-g/|\lambda_1|$. A small relative gap gives slow convergence.<Templates for the Solution of Algebraic Eigenvalue Problems: A Practical Guide/Power Method>{ref}
The short proof is the exact <eigenbasis> identity
$$
\frac{A^kx_0}{c_1\lambda_1^k}=v_1+\sum_{j=2}^n\frac{c_j}{c_1}\left(\frac{\lambda_j}{\lambda_1}\right)^kv_j.
$$
The remaining sum has <norm> bounded by a constant times $\gamma^k$. This proves convergence of direction, allowing a changing sign or complex factor. A tiny $c_1$ or a poorly conditioned, <nonorthogonal> <eigenbasis> can delay visible convergence: the <condition number> of $V$ affects constants, while $\gamma$ controls the eventual rate. The bound need not be an equality; if the initial coefficient of the second-largest <eigenvalue> vanishes, the <active spectrum of the power method> gives a faster rate. When $\gamma=0$, all unwanted components disappear after one multiplication.<Numerical Analysis, Spring 2022/Power iteration (2022-03-07 lecture notes)>[§1, PDF pp. 1–2: diagonalizable expansion and nonzero first coordinate of the starting vector]{ref}
The hypothesis is $c_1=(V^{-1}x_0)_1\ne0$, not an ordinary nonzero <inner product> with the right <eigenvector>. Let $w_1^*$ be the first row of $V^{-1}$, a <left eigenvector> normalized so $w_1^*v_j$ is $1$ for $j=1$ and $0$ otherwise; $*$ denotes <conjugate transpose>. Then $c_1=w_1^*x_0$. For a <Hermitian matrix>, choose an <orthonormal eigenbasis> and the criterion becomes $v_1^*x_0\ne0$: the starting <vector> must be <nonorthogonal> to the dominant <eigenvector>. For a general <diagonalizable matrix>, the <nonorthogonal> <eigenbasis> makes these two tests different, as illustrated under <power iteration failure cases>.
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