Suppose $H$ is an $n\times n$ symmetric positive definite matrix, $M_k$ is a sequence of $n \times n$ matrix (not necessarily symmetric) such that $M_k \to O$ where $O$ is the zero matrix. Let $\lambda_i(H),i=1,...,n$ denote the operator that gets the $i$th largest eigenvalue of $H$ in absolute value. My question is, is it true that $\mathop {\lim }\limits_{k \to \infty } \lambda_i (H + {M_k}) = \lambda_i (H),i=1,...,n$?

If this is not true for every $i$, is it true for $i=1$ (largest eigenvalue) and $i=n$ (smallest eigenvalue)? (My application only needs this one to hold)

Any explanation, counterexample or reference is helpful. Thanks!

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    $\begingroup$ Isn't this straightforward? Just diagonalize $H$, and the corresponding perturbed sequence converges to a diagonal matrix ...; this seems to only require that $H$ be diagonalizable. $\endgroup$ – David Handelman Apr 7 '18 at 3:00

As stated, $H+M_k$ need not have real eigenvalues; what is the largest then. If $f_A(\lambda)$ is the characteristic polynomial of a matrix $A$, then a simple eigenvalue depends locally real analytically on $A$, by the implicit function theorem.

If all eigenvalues are real and you order them by size, then each eigenvalue depends continuously on $A$, even if they cross eachother. For information on this and more general situations see this paper and references therein.


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