Do Approximate Eigenvalues Imply Approximate Eigenvectors?

Question

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!

Robert Israel · Answer

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.

copper.hat · Answer

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 $A v = lambda v$.

Do Approximate Eigenvalues Imply Approximate Eigenvectors?

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