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!

Having $\|A - B\|$ small is not enough, in itself, to make
$u_i$ and $v_i$ close.  Consider any real symmetric matrices $A_0$ and $B_0$ (with distinct eigenvalues to avoid any problems of degeneracy), and take $A = t A_0$ and $B = t B_0$.  Thus
$\|A - B\| = |t| \|A_0 - B_0\|$ can be made arbitrarily small by taking $t$ to be small.  But the eigenvectors of $A$ and $B$ are the same as the eigenvectors of $A_0$ and $B_0$, and thus don't have to be close at all.
