Problem:
Let $A,~E \in M_{n\times n}(\mathbb{R})$ be two $n\times n$ symmetric matrices over $\mathbb R$ and $q$ be an unit eigenvector of $A$. Here $A$ is what we are care about and $E$ is a disturbance, and we are going to research about eigenvector of $A+E$.
WLOG, let $A=\begin{bmatrix}\lambda&0\\0&D_2\end{bmatrix}$ and $E=\begin{bmatrix}\varepsilon&e^T\\e&E_2\end{bmatrix}$， where $D_2,E_2$ are $(n-1)\times (n-1)$ symmetric matrices and $e \in \mathbb R^n$, then $q=e_1$. Let $d:=\min\limits_{\mu \in \lambda(D_2)}|\lambda-\mu|$, where $\lambda(D_2)$ is the set of all eigenvalue of $D_2$, then we have following theorem:
if $d>0$ and $\Vert{E}\Vert_2\le \frac{d}{4}$, then there exists an unit eigenvector $v$ of $A+E$ such that $\sin\theta:=\sqrt{1-|q^Tv|^2}\le \frac{4}{d}\Vert{e}\Vert_2\leq \frac{4}{d}\Vert{E}\Vert_2$
My current achievements:
Let $v=(v_1,w^T)^T,~w \in \mathbb R^{n-1}$. Then if $(A+E)v=\rho v$, we have $$ \begin{align} e^Tw&=(\rho-\lambda-\varepsilon)\\ v_1e&=(\rho I-D_2-E_2)w \end{align} $$ Since $v$ is unit vector, $\sin\theta=\sqrt{1-v_1^2}=\Vert{w}\Vert_2$, so we just need to prove that $$ \frac{\Vert e\Vert_2}{\Vert w\Vert_2}\ge \frac{d}{4} $$ After that, I've tried a lot but nothing works. Can someone please give me some hint?