If I take a set of normalized eigenvectors $h_n$ (in $\mathbb R^n$) that correspond to the largest eigenvalues of matrices $A_n$, is this set of eigenvectors compact - i.e., closed and bounded?
The set is obviously bounded, but I'm not sure whether the set is closed - I am probably just overlooking something trivial.
Do I just consider a convergent sequence in this set and since the limit vector will again be in the set (normalized, corresponds to the largest eigenvalue of some matrix $A_n$) then by definition this set is closed?
Thanks,
The set is finite (it has no more than $n$ linearly independent vectors), so it's trivially compact.
If you would like to see this with a sequence, take any infinite sequence $\{x_n\}_n$in that set. As the set is finite, at least one of the elements of the set must be infinitely repeated in the sequence, let's call it $y$. Then, the constant subsequence $\{y\}_k=\{x_{n_k}\}_k$ is obviously convergent, then the set is compact.
Note: I never used that we're working in $\Bbb R^n$, in fact, this proof works for any finite subset of a normed space.