Convergence of eigenvectors Let $\Sigma$ be a $k\times k$ positive definite matrix and let $V$ be the diagonal matrix containing its eigenvalues, which we assume to be distinct.
Suppose $H_n$ is a sequence of $k\times k$ matrices such that
$(i)$ $\Sigma H_n \to H_nV$ as $n\to \infty$
$(ii)$ $H_n^{'}H_n \to I_k$ as $n\to \infty$
with respect to the Frobenius norm. How can I show that $H_n$ converges to the orthonormal eigenvectors of $\Sigma$? Any help is greatly appreciated.
 A: Proof that the columns converge up to a sign change:
First, by considering the diagonal entries of $H_n'H_n$, conclude that all columns have lengths that converge to $1$.
Now, let $v_n$ denote the $j$th column of $H_n$, and let $\lambda$ denote the $j$th diagonal entry of $V$ (i.e. the $j$th eigenvalue).  Let $v_*$ denote a known eigenvector of $\Sigma$ associated with $\lambda$.
By condition (i), we can conclude that $\lim_{n \to \infty} \Sigma v_n = \lambda v_n$.  In other words, $\lim_{n \to \infty}(\Sigma - \lambda I_k)v_n = 0$.  Since $\|v_n\| \to 1$, we can conclude that $(v_n)_{n \in \Bbb N}$ forms a bounded sequence.  If $v_n$ converges then there's nothing to do, so we assume that it does not converge.
Now, suppose that $w$ is a cluster point of the sequence $v_n$.  In other words, there exists a subsequence $v_{n_p}$ such that $\lim_{p \to \infty} v_{n_p} = w$.  Because the functions $x \mapsto \|x\|$ and $x \mapsto (\Sigma - \lambda I_k)x$ are continuous, we have
$$
\|w\| = \left\| \lim_{p\to \infty} v_{n_p}\right\| = \lim_{p\to \infty} \|v_{n_p}\| = 1,\\
(\Sigma - I_k)w = (\Sigma - I_k)\lim_{p\to \infty} v_{n_p} = 
\lim_{p\to \infty} (\Sigma - I_k) v_{n_p} = 0.
$$
Since $w$ satisfies both of these, we must have $w = v_*$ or $w = -v_*$.
Now, we need a tricky analysis statement: because $(v_n)$ is a non-convergent bounded sequence in $\Bbb R^n$ with cluster points $v_*$ and $-v_*$, $(v_n)$ can necessarily be broken into two subsequences, one of which converges to each limit point.  
In other words, there exists a set $S_+ \subset \Bbb N$ and $S_- = \Bbb N \setminus S_+$ such that $(v_n)_{n \in S_+}$ converges to $v_*$ and $(v_n)_{n \in S_-}$ converges to $-v_*$.  To put that another way, if we define
$$
d_n = \begin{cases}
1 & n \in S_+\\
-1 & n \in S_-,
\end{cases}
$$
then we find that the sequence $d_nv_n$ converges to $v_*$.

Let the sequence $d_n^{(j)}$ be defined as follows. $d_n = \pm 1$ for all $n$ if the $j$th column sequence $v^{(j)}_n \to \pm v_*^{(j)}$.  If $v_n^{(j)}$ fails to converge, then define $d_n^{(j)}$ to be the sequence $d_n$ constructed above.  It then suffices to take $d_n^{(j)}$ to be the $j$th diagonal entry of $D_n$.
So $H_n$ indeed converges "up to sign change".

We would need, first of all, the assumption that $H_n$ converges at all. 
If we know that $H_n$ converges, then continuity does the rest of the work.  In particular, the functions $X \mapsto AX$, $X \mapsto XB$, and $f:X \mapsto X'X$ are all continuous.
It follows that
$$
\Sigma H = \Sigma \lim_{n \to \infty}H_n = \lim_{n \to \infty} \Sigma H_n = \lim_{n \to \infty} H_n V = \left(\lim_{n \to \infty} H_n \right)V = HV,\\
H'H = f(H) = f\left(\lim_{n \to \infty} H_n \right) = \lim_{n \to \infty }f(H) = \lim_{n \to \infty}H_n'H_n = I_k.
$$
That is, $\Sigma H = HV$ and $H'H = I_k$. It follows that $H$ is an orthogonal matrix whose columns are eigenvectors of $\Sigma$, as desired.
