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!



Limit of eigenvalues of a matrix perturbation sequence
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](http://www.mat.univie.ac.at/~michor/DC-perturb.pdf) and references therein. 


