Eigenvector convergence Suppose $\{A_k\}, A$ are $\mathbb{R}^{n \times n}$ matrices such that $A_k \rightarrow A$. How do I show that the dominant eigenvectors of $A_k$ converge to the dominant eigenvector of $A$. By dominant eigenvector, I mean eigenvector corresponding to the eigenvalue with the maximum absolute value. 
 A: Suppose $A_k x_k = \lambda_k x_k$, with $\|x_k\|=1$, $\lambda_k$ being one dominant eigenvalue of $A_k$, which means that 
$A_k - \tau I$ is invertible for all $\tau$ with $|\tau|>|\lambda_k|$.
Then by extracting a subsequence, $x_{k'}\to x$ with $\|x\|=1$. This implies
that 
$$
\lambda_{k'} x_{k'} = A_{k'} x_{k'} \to Ax.
$$
Since $(x_k)$ is bounded, it follows that $(\lambda_k)$ is bounded as well. Thus, we can extract another subsequence such that $\lambda_{k''}\to \lambda$. Then it follows
$$
Ax = \lambda x,
$$
and since $\|x\|=1$, this implies that $(x,\lambda)$ is an eigenpair.
It remains to prove that $\lambda$ is the dominant eigenvalue of $A$, i.e. $|\lambda|>|\lambda'|$ for all other eigenvalues of $A$. Let me show this under the assumption that all the matrices $A_k$ is symmetric (or complex normal $AA^H=A^HA$). I do not know how to prove this in the general case.
For such matrices, it holds $|\lambda_k|=\|A_k\|_2$, where $\|\cdot\|_2$ is the matrix 2-norm. 
Since norms on finite-dimensional spaces are equivalent it follows $\|A_k\|_2\to \|A\|_2$.
Passing to the limit on the subsequence, yields $|\lambda|=\|A\|_2$.
Since $A$ is symmetric (or normal), $|\lambda|$ is the dominant eigenvalue.
Hence, we proved that any subsequence of $(x_k,\lambda_k)$ contains a subsequence converging to a $(x,\lambda)$, where $\lambda$ is the dominant eigenvalue of $A$ and $x$ is some eigenvector to $\lambda$.
It follows that the sequence $(\lambda_k)$ converges to $\lambda$. Convergence of eigenvector cannot be expected without some further normalization ($x_k$ and $-x_k$ both satisfy the assumptions, so we can multiply some of the vectors $x_k$ by $-1$ without violating the assumptions).
