Do Approximate Eigenvalues Imply Approximate Eigenvectors? My apologies in advance if this has already been asked somewhere.
Suppose I have two real symmetric matrices $A$ and $B$ in $\mathbb{R}^{d \times d}$ for which $\lVert A - B \rVert_{op} \le \varepsilon$. Further, call the eigenvalue-eigenvector pairs for $A$ and $B$ as $(\lambda_i, u_i)$ and $(\tau_i, v_i)$, for all $i \in [d]$, and suppose that $\lVert u_i \rVert_2 = \lVert v_i \rVert_2 = 1$ for all $i \in [d]$.
My question is: under what condition can we say something interesting about $\lVert u_i - v_i \rVert_2$?
So far, I've tried using the following facts.

*

*For all $i$, $\lvert \lambda_i - \tau_i \rvert \le \varepsilon$.

*If $\lvert \lambda_i - \tau_i \rvert \le \varepsilon$, then we can write $\lVert Bu_i - \lambda_i u_i \rVert \le \varepsilon$ (the reason I thought this might be useful is that it shows that the eigenvalue-eigenvector pairs for $A$ are almost eigenvalue-eigenvector pairs for $B$, in some sense)

I'm not sure where to go from here, or if I should be looking someplace else entirely.
Thank you in advance for the help!

Eigenvalues are continuous in a certain prescribed sense, but eigenspaces may shrink.
Let $A(\epsilon) = \begin{bmatrix} 0 & 0 \\ 0 & \epsilon \end{bmatrix}$.
Note that for $ \epsilon \neq 0$ that $A(\epsilon)$ has two eigenvectors $(1,0)^T, ( {1 \over \epsilon} , 1)^T$ but for $\epsilon = 0$ any point is an eigenvector.
Hence $(1,1)^T$ is an eigenvector of $A(0)$ but not of any $A(\epsilon )$ for $\epsilon \neq 0$.
However, they are 'continuous' in the sense that if $v_n$ are unit vectors such that $v_n \in \ker (A_n-\lambda_n I)$, $A_n \to A$ then there is some subsequence $K$ such that $\lambda_n \xrightarrow{K} \lambda$, $v_n \xrightarrow{K} v$ and $A v = \lambda v$.
