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 $i$th largest eigenvalue of $H$. 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!
Your problem is not well-posed. Indeed the eigenvalues of $H+M_k$ are not necessarily real. What is $\lambda_i(H+M_k)$ ?
Roughly speaking, the result is true for any matrices (cf. the Berci's comment). More precisely,
Let $(A_k)_k$ be a sequence of matrices $\in M_n(\mathbb{C})$ that converges to $A\in M_n(\mathbb{C})$. Then there are orderings of $spectrum(A_k)=(\lambda_{k,i})$ and of $spectrum(A)=(\lambda_i)$ s.t., for every $i$, $(\lambda_{k,i})_k$ tends to $\lambda_i$.