Let $A$ be an $n\times n$ matrix of all ones. Consider the analytic perturbation of $A$ as $$\tilde{A} = A + \epsilon H\_1 + \epsilon^2 H\_2 + \epsilon^3 H\_3 + ... $$ All matrices are symmetric. Assume $\tilde{A}$ to be positive definite. Let $E = [e\_0,e\_1,e\_2,...e\_{n-1}]$ be the eigenvectors matrix of $A$ and $\tilde{E} = [\tilde{e}\_0,\tilde{e}\_1,\tilde{e}\_2,...\tilde{e}\_{n-1}]$ be the eigenvectors matrix of the matrix $\tilde{A}$.


I have seen David Kahan theorem bounds the sine of the angle between the eigenvectors stating $\lim\limits\_{\epsilon\to 0}\sin\Theta(e\_i,\tilde{e}\_i) = 0$ and also another theorem which says $\lim\limits\_{\epsilon\to 0}\|E-\tilde{E}\|\_{\infty} = 0$ or to this effect in this [paper](https://arxiv.org/pdf/1603.03516.pdf) and also this [paper](http://proceedings.mlr.press/v83/eldridge18a/eldridge18a.pdf). I am sure from these references that these statements hold when eigenvalues of $A$ are distinct.


I have seen some vague references but not sure if the above statements hold when there are eigenvalues of multiplicity greater than 1. (It is the case here, as eigenvalues of $A$ are $n$ with multiplicity $1$ and $0$ with multiplicity $n-1$.


So in this case can we say that $\lim\limits\_{\epsilon\to 0} \|\tilde{E}-E'\|\_{\infty} = 0$ where $E' = E$ up to a permutation of columns.


PS : I myself am not doing research in perturbation theory, but this situation arise in my work and I referred to these papers to see if the eigenvectors matrix converges. The message in those papers are a bit cryptic and was not sure if these papers apply in this case where there are eigenvalues of multiplicity higher than $1$.



Convergence of the eigenvector matrix for an analytic perturbation of a singular matrix
Because of the degeneracy in the eigenvalues of $A$, "the eigenvectors matrix of $A$" is far from well-defined. Rather, there is a one-dimensional eigenspace of $A$ for eigenvalue $n$ (spanned by $(1,\ldots,1)^T$), and its orthogonal complement is the eigenspace for eigenvalue $0$. For any $\eta > 0$, there is $\delta > 0$ such that
if $|\epsilon| < \delta$, $\tilde{A}$ has one eigenvalue within distance $\eta$ of $1$, all the others within distance $\eta$ of $0$, and the orthogonal projections for $A$ and $\tilde{A}$ on the span of eigenspaces for eigenvalues within $\eta$ of $0$ (or $1$) are within distance $\eta$ (for whichever matrix norm you prefer).


