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.
Best Answer
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.