When do eigenvectors converge?

Question

Let $A_n$ be a sequence of self-adjoint $N\times N$ matrices that converge in the operator norm to $A$. The sequence of eigenvalues of $A_n$, denoted $\lambda_n$, converges to an eigenvalue of $A$, denoted $\lambda$.

How do you show that the sequence of eigenvectors of $A_n$ corresponding to eigenvalue $\lambda_n$ converges to the eigenvector of $A$ corresponding to eigenvalue $\lambda$ (in the usual vector norm)?

EDIT: sorry guys, I meant to say that all matrices have non degenerate eigenvalues.

Answer 1

If $A_n$ converges to $A$ in norm, $\lambda_n$ is an eigenvalue of $A_n$ with eigenvector $v_n$. If $\lambda_n\to\lambda$ and $v_n\to v\ne 0$, then $\lambda$ is an eigenvalue of $A$ with eigenvector $v$.

For all $n$ it holds: $$ Av-\lambda v = A(v-v_n) + (A- A_n)v_n+ (A_nv_n - \lambda_n v_n) + (\lambda_n-\lambda)v_n+\lambda (v_n - v) . $$ The third addend is zero, all other addends tend to zero for $n\to\infty$, hence $$ Av-\lambda v =0. $$

Answer 2

Daw, if $v_n$ tends to $v\not=0$, then of course $Av=\lambda v$ ; yet, that is not the question !

Let $A_n=\begin{pmatrix}1/n&\sin(n)/n\\\sin(n)/n&-1/n\end{pmatrix}$. The matrices $A_n$ are real symmetric and tend to $0$. Consider the positive eigenvalue of $A_n$: $\lambda_n=\sqrt{1/n²+(\dfrac{\sin(n)}{n})²}$. An associated eigenvector is $v_n=[1,\dfrac{n\lambda_n-1}{\sin(n)}]^T=[1,\dfrac{\sin(n)}{\sqrt{1+{\sin(n)}²}+1}]^T=[1,y_n]^T$. Since $y_n$ and $1/y_n$ have no limit, then we cannot choose a sequence $(v_n)_n$ of eigenvectors of $(A_n)_n$ associated to $(\lambda_n)_n$ that converges. The problem is that the limit matrix has a double eigenvalue. We have not this pathology when the limit matrix has distinct eigenvalues.