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Suppose $A$ is a positive semi-definite, Hermitian matrix with a unit-norm eigen vector $\textbf{v}$ corresponding to its largest eigen value $\lambda$. Let $B = A + \alpha \textbf{z}\textbf{z}^H$, where $\alpha \ge 0$ and $||\textbf{z}||_2=1$. If $\textbf{w}$ is a unit-norm eigen vector of $B$ corresponding to its largest eigen value, can we upper bound the distance between dominant eigen spaces of $A$ and $B$ given by $d=||\textbf{w} - c\textbf{v}||_2$, where $c = \textbf{v}^H\textbf{w}$?

Clearly $d \le 2$, but I wish to see a better bound in terms of $A, \alpha, \textbf{v}$ and $\textbf{z}$.

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A tiny change can result in a shift of the dominant eigenvalue's eigenspace to something that is completely orthogonal to the dominant eigenvalue's eigenspace of $A$.

For example, let

$A=I-\epsilon \left[ \begin{array}{cccc} 1 & 0 & \ldots & 0 \\ 0 & 0 & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \ldots & 0 \end{array} \right] $

Let $z=\left[ \begin{array}{c} 1 \\ 0 \\ \vdots \\ 0 \end{array} \right] $

and $\alpha=2\epsilon$.

The general issue here is that within repeated eigenvalues, the eigenvectors aren't continuous with respect to changes in the matrix.

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  • $\begingroup$ Thanks. A small change, in the example provided, can lead to a distance of $d = 1$ between dominant eigenvalues's eigenspace. But consider building a Hermitian matrix $A_{nxn}$ through sequential addition of rank-one Hermitian matrices as in: $A(k) = A(k-1)+\alpha_k \textbf{a}_k\textbf{a}_k^H, k=1,2,...$, where $A(0)=0$ and $\alpha_1 \ge ... \ge \alpha_k > 0$. In this case eigenvector corresponding to largest eigenvalue evolves as $\textbf{v}_k, k=1,2...$, starting with $\textbf{v}_1=\textbf{a}_1$. Can we bound $||\textbf{v}_k-\textbf{v}_{k-1}||_2, k>1$? $\endgroup$
    – Arun
    Commented Nov 18, 2013 at 3:36

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