Suppose we are given two positive semidefinite matrices $A$ and $B$ that satisfy $Tr(A)=Tr(B)=1$ and $\|A-B\|_1 \leq \varepsilon$. We also assume that largest eigenvalue of $A$ is $\delta$ and rest other eigenvalues are smaller than $\delta^2$. This means that there is a large gap between the largest eigenvalue and second largest eigenvalue of $A$. Further, suppose that $\varepsilon << \delta$, for example $\varepsilon= \delta^2$. Then what can be said about the closeness between the largest eigenvector of $A$ and largest eigenvector of $B$? Here, the word `largest eigenvector' implies the eigenvector corresponding to the largest eigenvalue.
We know from Weyl's inequality that the largest eigenvalues of $A$ (call it $\lambda_{max}(A)$) and $B$ (call it $\lambda_{max}(B)$) satisfy $|\lambda_{max}(A)-\lambda_{max}(B)|\leq \varepsilon$. My question asks if similar statement is true for largest eigenvectors as well, given the aforementioned assumption of large eigenvalue gap in $A$.
By Davis-Kahan's sin-Theta theorem $$ \|v_1(A)-v_1(B)\|_2 \leq \frac{\sqrt{2}\|A-B\|_2}{\lambda_1(A)-\lambda_2(B)}. $$