Let $A_n$ be a sequence of $d\times d$ symmetric matrices, let $A$ be a $d\times d$ symmetric positive definite matrix (matrix entries are assumed to be real numbers). Assume that each element of $A_n$ converges to the corresponding element of $A$ as $n\to \infty$. Can we conclude that for some $\epsilon>0$ there exists $N_\epsilon$ such that the smallest eigenvalue of $A_n$ is larger than $\epsilon$, for all $n \geq n_\epsilon$?
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Does this answer your question? Eigenvalues of matrix with entries that are continuous functions – Eric Wofsey Aug 08 '21 at 21:47
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Or maybe a better duplicate target: https://math.stackexchange.com/questions/1346866/are-eigenvalues-of-the-limit-of-a-sequence-of-matrices-limits-of-eigenvalue-sequ – Eric Wofsey Aug 08 '21 at 21:49
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@Eric Wofsey I think that the answer to the second question you've pointed out does reply to mine too, thanks. However, I quite like Yorch's answer (which is given from a "norm perspective"). – Jack London Aug 09 '21 at 08:04
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Yes, since the coefficients of the characteristic polynomials of the $A_n$ converge to those of the char. polynomial of $A$, and the roots of a polynomial depend continuously on the coefficients.
Hans Engler
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Let $\lambda_1 \geq \dots \lambda_n$ be the eigenvalues. Any $e< \lambda_n$ works.
Suppose $\lambda < e$ is an eigenvalue of $A_n$, take an orthonormal basis of $A$, take a unitary eigenvector of $A_n$ for $\lambda$ and let the coordinates in the basis be $(a_1,\dots,a_n)$.
Note $A_nv = (\lambda a_1,\dots, \lambda a_n)$ and $Av = (\lambda_1a_1, \dots,\lambda_n a_n)$ so that the distance between the two vectors is $\sqrt{ \sum \limits_{i=1}^n (\lambda_i - \lambda)^2a_i^2 } \geq \lambda_n - \lambda$
By unicity of norms we have that $A_n$ converges to $A$ under the usual norm which is the supremum of the distances between $Av$ and $A_nv$ for $v$ in the unitary sphere. It follows that $A_n$ cannot have an eigenvalue less than $e$ for sufficiently large $n$.
Asinomás
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