Let $R=S+xx^t$
where $x\in \mathbb{R}^n$ and $R$ and $S$ are $n\times n$ symmetric positive semidefinite matrices.
Is there anything I can say about the difference of the first eigenvectors of $R$ and $S$ in terms of $x$? In this question the relation between eigenvalues are discussed well, but my quesion is about eigenvectors.
Let $W_R$ and $W_S$ be the eigenvectors of $R$ and $S$ related to their largest eigenvalues $\lambda_R^{max}$ and $\lambda_S^{max}$, respectively. I would like to show that $\vert\vert{W_R-W_S}\vert\vert_2$ is related to $\vert\vert{x}\vert\vert_2$, such as: $$\lim_{\vert\vert{x}\vert\vert\to 0}\vert\vert{W_R-W_S}\vert\vert \to 0$$
Is there any suggestion?
Without loss of generality, take all our eigenvectors to be unit vectors.
Note that $W_R$ is the unit vector (actually, one of at least two unit vectors) that maximizes the function $$ f_R(y) = y^TRy $$ whereas $W_S$ is the unit vector that maximizes the function $$ f_S(y) = y^TSy = f_R(y) + (y^Tx)^2 $$ Intuitively, we'd like to be able to say that if $x$ is small enough, then the eigenvector shouldn't change that much. In fact, we should say that $x$ is small relative to the difference between the biggest two eigenvectors.
I will now assume that $R$ has only a one-dimensional eigenspace for its largest vector. Otherwise, the question is going to need a lot of reframing. Let $a$ denote the largest eigenvalue of $R$, and let $b$ denote the second-largest. Consider (an arbitrarily small) $\epsilon > 0$.
Suppose that $\|x\|/(a-b) < \epsilon$. We then note that $$ f_S(W_R) = f_R(W_R) + (y^Tx)^2 \geq b $$ Moreover, and this is key, note that for $y$ that are sufficiently close to $W_R$, we have (using unitary diagonalizability) $$ f_R(W_R) - f_R(y) \geq (b-a)\|W_R - y\|^2 $$ Honestly, I'm not sure where to go from here right now, but I'm hoping that this is a helpful start.