TITLE: Are eigenvalues of the limit of a sequence of matrices limits of eigenvalue sequences?
QUESTION [16 upvotes]: Let $\{A_n\}\in \mathbb{R}^{m\times m}$ be a sequence of symmetric matrices such that $A_n\to A$ as $n\to \infty$, i.e. $\lim_{n\to \infty}a_{ij}(n)=a_{ij}\ \forall 1\le i,j\le m$ where $A_n=[a_{ij}(n)],A=[a_{ij}]$. Let $\rho(A_n)=\{\lambda_1(n),\cdots,\ \lambda_m(n)\}$ be the eigenvalues of $A_n$ and similarly, $\rho(A)=\{\lambda_1,\cdots,\ \lambda_m\}$ be the eigenvalues of $A$, arranged in, say, increasing order. Here are my questions

1)Can I write $\lambda_k(n)\to \lambda_k,\ 1\le k\le m$?
2)If I define (with a slight abuse of standard notation) $\delta_s(n),\ 1\le s\le m$ as the maximum eigenvalue of any $s\times s$ submatrix of $A_n$ and if $\delta_s$ be the corresponding quantity for $A$, then can I say that $$\lim_{n\to \infty}\delta_s(n)=\delta_s$$ ?

Intuitively it seems to me that the answers are positive since the eigenvalues of a matrix are continuous functions of the elements of the matrix and $\delta_s$ is just the maximum of some eigenvalues of submatrices.
However, I am not sure if this argument is sound enough. Maybe this is a very trivial issue for people here, but I would really appreciate if someone can kindly provide some explanation regarding the correct answer. Thanks in advance.

REPLY [3 votes]: The answer of @Hippalectryon is morally correct. However, there is this problem of approaching a polynomial with multiple roots, somehow delicate. $\tiny{\text{Can be proved with the argument principle from complex analysis.}}$ We'll work directly with the eigenvalues, bypassing the characteristic polynomial.
Use this standard observation: if two symmetric matrices are ordered 
$B\prec C$ then their vectors of eigenvalues ( ordered increasingly) 
also satisfy $\lambda(B) \prec \lambda(C)$ ( that is $\lambda_i(B) \le \lambda_i(C)$ for all $i$). 
Now, if if $A_n \to A$ then for any $\epsilon >0$ we have $A- \epsilon I \prec A_n \prec A + \epsilon I$  for $n \ge n_{\epsilon}$, and so 
$$\lambda_i(A) - \epsilon \le  \lambda_i(A_n) \le \lambda_i(A) + \epsilon$$
$\tiny{\text{(the question for the max for submatrices is simple: $\lim$ and $\max$ commute ).}}$
$\bf{Added:}$ I am giving this " continuity of the roots" question more thought. Of course, @Hippalectryon is right. But how do we truly convince ourselves that the roots of $P_n$ approach the roots of $P$? First, a hands-on approach: The roots of $P_n$ will all lie in a bounded region since the coefficients of $P_n$ are all bounded. If the roots of $P_n$ did not approach those of $P$ then we would find, by compactness, a subsequence whose roots approach some other $n$-uple. But that would mean in the limit that $P$ would decompose using that $n$-uple, that is, it would have two distinct decompositions,  contradiction. 
The high-brow explanation is that the map roots $\mapsto$ polynomials from $\mathbb{C}^n$ to $\mathbb{C}^n$ is continuous and proper and induces a bijective map $\mathbb{C}^n/S_n \to\mathbb{C}^n$, which is bijective, continuous and closed, hence a homeomorphism.  $\tiny{\text{(surjectivity is equivalent to the fundamental theorem of algebra)}}$