Nearby matrices have nearby leading eigenvectors? Suppose I have a symmetric positive semidefinite matrix $A$ with leading eigenvalue $1$ of multiplicity $1$ and remaining eigenvalues $\leq\epsilon$. I am told that another symmetric positive semidefinite matrix $B$ is close in some sense to $A$. I wish to conclude that the leading eigenvector of $B$ is therefore close in some sense to that of $A$.
Question: What notions of closeness correspond to existing theorems of this sort?
One solution: Let $a$ and $b$ denote leading eigenvectors of $d\times d$ matrices $A$ and $B$, respectively, scaled so that $\|a\|_2^2=1$ and $\|b\|_2^2$ is the leading eigenvalue of $B$. Then by triangle, Eckart–Young–Mirsky, and triangle again, we have
$$\|bb^\top-aa^\top\|_F\leq\|bb^\top-B\|_F+\|B-aa^\top\|_F\leq2\|B-aa^\top\|_F\leq2\big(\|B-A\|_F+\|A-aa^\top\|_F\big)\leq2\|B-A\|_F+2\sqrt{d-1}\epsilon.$$
For this attempt, I don't like the additive loss in $\epsilon$, and I doubt it's necessary. Can I get better performance with a different norm? Perhaps the spectral norm?
 A: You can get a nice estimate using the spectral norm.
Theorem 3.2.32 of Bratteli and Robinson vol. 1 tell us an estimate on $f(A) - f(B)$ for a variety of functions.  Take $f:\mathbb R \rightarrow \mathbb R$ to be zero on  $(-\infty, \epsilon]$ and one on $[1,\infty)$, interpolating with some quadratics.  I think you get
$$
\|f(A) - f(B)\| \leq  \frac{2\|A - B\|}{1-\epsilon} 
$$
and this helps.  Notice $f(A) = aa^*$  and  $f(B) = bb^*$ and I assume you used transpose because you assumed real scalars.  Works either way, I think.
I should add I am not entirely sure of the constant, just that the bound in linear in $\|A - B\|$.
A: The spectral projection of $A$ for eigenvalue $1$ can be realized as
$P_A = \dfrac{1}{2\pi i}  \oint_{\Gamma} (z I - A)^{-1} \; dz$
where $\Gamma$ is a contour that encloses $1$ but none of the other 
eigenvalues.  In particular, take $\Gamma$ to be a circle
of radius $r$ centred at $1$, where $r = (1-\epsilon)/2$.  For $z \in \Gamma$ we have
(using the $\ell^2$ operator norm)
$\|(z I - A)^{-1}\| \le 1/r$.  Then if $\|B - A\| < r$, 
$z I - B$ is invertible for $z \in \Gamma$, with 
$$\|(z I - B)^{-1} - (zI-A)^{-1} \| \le \dfrac{\|(z I - A)^{-1}\|^2 \|B-A\|}{1 - \|B - A\| \|(zI - A)^{-1}\|} \le \dfrac{\|B-A\|}{r(r - \|B-A\|)}$$
and $P_B = \dfrac{1}{2\pi i} \oint_{\Gamma}  (z I - B)^{-1} \; dz$
is a spectral projection of rank $1$ 
with 
$\|P_B - P_A\| \le \dfrac{\|B-A\|}{r - \|B-A\|}$.
