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.

1. For all $i$, $\lvert \lambda_i - \tau_i \rvert \le \varepsilon$.
2. 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!

Answer 1

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 $A v = \lambda v$.

Answer 2

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.