Why does the power method converge to an eigenvector? I'm currently working on a way to find eigenvectors analytically. The teacher suggested that I show that
$$ x_{k+1} \gets \dfrac{Ax_k}{\|Ax_k\|}$$
where $x_{k}$ is the vector and $A$ the matrix, converges to an eigenvector.
I can't seem to think of any ideas on how to prove this. Is there any literature on this?
 A: This method is known as the power iteration, you can first check that without paying attention to $0$ division, $x_n = \frac{A^nx_0}{||A^nx_0||}$.
To see what's happening let me check this convergence when A is symmetric positive and has $d$ distinct eigenvalues and $x_0$ has a non negative component with respect to the eigenvector of largest magnitude. In this case one can introduce an orthogonal basis $(v_1,...,v_d)$ of $\mathbb{R}^d$ constituted of eigenvectors associated to $A$'s eigenvalues $0 \leq \lambda_1 < ... < \lambda_{d-1} < \lambda_d$. Then by Pythagoras's formula we have for any $x \in \mathbb{R}^d$ and $n \in \mathbb{N}$,
$$\Vert A^nx\Vert^2 = \left\Vert A^n\sum_{k=1}^d \langle v_k,x \rangle v_k\right\Vert^2 = \left\Vert\sum_{k=1}^d \langle v_k,x \rangle\lambda_k^nv_k\right\Vert^2 = \sum_{k=1}^d \langle v_k,x \rangle^2\lambda_k^{2n} $$
and thus, using that $\langle v_k,x_0 \rangle > 0$,
$$\frac{A^nx_0}{\Vert A^nx_0\Vert} = \sum_{k=1}^d \frac{\langle v_k,x_0 \rangle\lambda_k^n}{\sqrt{\sum_{k=1}^d \langle v_k,x_0 \rangle^2\lambda_k^{2n}}}v_k$$
Since $(v_1,...,v_d)$ is a basis of $\mathbb{R}^d$ we only have to check each real sequence of coordinates, if $k < d$ we have
$$\frac{\langle v_k,x_0 \rangle\lambda_k^n}{\sqrt{\sum_{k=1}^d \langle v_k,x_0 \rangle^2\lambda_k^{2n}}} = \frac{\langle v_k,x_0 \rangle(\lambda_k / \lambda_d)^n}{\sqrt{\sum_{k=1}^d \langle v_k,x_0 \rangle^2(\lambda_k / \lambda_d)^{2n}}} = \frac{o(1)}{\sqrt{\langle v_k,x_0 \rangle^2 + o(1)}} \longrightarrow 0$$
and for $k = d$,
$$\frac{\langle v_d,x_0 \rangle\lambda_d^n}{\sqrt{\sum_{k=1}^d \langle v_k,x_0 \rangle^2\lambda_k^{2n}}} = \frac{\langle v_d,x_0 \rangle}{\sqrt{\langle v_k,x_0 \rangle^2 + o(1)}} \longrightarrow 1$$
This can be adapted to the case of any number of complex eigenvalues but constants may appear.
