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.

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    $\begingroup$ You need to be careful here; there's the possibility that the eigenvalues may "switch". For example, we could have $\lambda_1(n) \to \lambda_2$ and $\lambda_2(n) \to \lambda_1$ $\endgroup$ Commented Jul 2, 2015 at 11:11
  • $\begingroup$ Outside of that technicality, your answer is indeed correct and sufficient. $\endgroup$ Commented Jul 2, 2015 at 11:13
  • $\begingroup$ If we have symmetric matrices (as you do now), the issue of "switching eigenvalues" isn't a problem for this question. $\endgroup$ Commented Jul 2, 2015 at 11:14
  • $\begingroup$ @Omnomnomnom, is the argument I provided correct? Also, can you kindly comment on the second question? $\endgroup$ Commented Jul 2, 2015 at 11:14
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    $\begingroup$ Yes, that's what I meant; your argument that the eigenvalues depend continuously on the matrix is enough to prove both 1) and 2). In order to get 2) quickly, we could note that the function $$ (x_1,\dots,x_n) \mapsto \max \{x_1,\dots,x_n\} $$ is continuous over $\Bbb R^n$. $\endgroup$ Commented Jul 2, 2015 at 11:17

2 Answers 2


Let $\chi_n$ be the characteristic polyomial of $A_n$ and $\chi=a\prod(X-\lambda_i)$ be that of $A$, then $\chi_n\mathop{\to}_{n\rightarrow\infty}\chi$.

If a sequence of polynomials converges to another, then by continuity the roots of the polynomials in the sequence must converge to the roots of the limit polynomial (with the same multiplicity). Hence the eigenvalues of the sequence of matrices indeed convergence to the eigenvalues of $A$.

Likewise, since any submatrix of $A_k$ converges to the corresponding submatrix of $A$, that is sufficient to say that $\delta_s(n)\to\delta_s$.

  • $\begingroup$ @orangeskid Thanks for the feedback, I added an explanatory line. $\endgroup$ Commented Jul 2, 2015 at 11:25
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    $\begingroup$ @Hippalectryon: It's tempting to pass to the characteristic polynomial and, since that varies continuously, use that the roots also do. That is correct but I would give it more thought. It's the problem with approaching a polynomial with multiple roots. Otherwise, the implicit function theorem works OK. $\endgroup$
    – orangeskid
    Commented Jul 2, 2015 at 11:30

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. $\tiny{\text{(surjectivity is equivalent to the fundamental theorem of algebra)}}$

  • $\begingroup$ By $A\prec B$ do you mean that the elements of $A$ are smaller than the elements of $B$, individually? $\endgroup$ Commented Jul 4, 2015 at 5:41
  • $\begingroup$ @Samrat Mukhopadhyay: I mean $B-A$ positive definite. $\endgroup$
    – orangeskid
    Commented Jul 4, 2015 at 8:30
  • $\begingroup$ Oh, ok, got it. $\endgroup$ Commented Jul 4, 2015 at 11:37

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