# eigenvector update formula

Suppose that $B$ is a Hermitian matrix with one known eigenpair $(\lambda,v)$. (assume its the smallest or largest pair, if you like). Form the rank one update $B+\rho bb^{T}$.

Now I'm interested in a (possibly approximate) formula for the updated eigenpair $(\widetilde{\lambda},\widetilde{v})$, that depends only on the entries of $B$ and $\lambda,v$. There are such formulas derived in a classic 1978 paper by Bunch, Nielsen and Sorensen (building on an earlier classic work by Golub), but they involve the full eigenbasis of $B$.

Any pointers?

P.S. (Assume $\lambda,\widetilde{\lambda}$ are simple, to keep things simple).

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By $A$ do you mean $B$? Edit: You will probably also want to know quantitatively how `simple' $\lambda$ is, for example something like the distance to its nearest eigenvalue neighbor. – tergi Aug 6 '12 at 23:24
I changed $A$ to $B$, thanks! I think I can obtain this information - what next? I eagerly await further enlightenment! :) – Felix Goldberg Aug 6 '12 at 23:50

Here is a first-order approximation. I'll write $\tilde{v} = v + \rho w + O(\rho^2)$ and $\tilde{\lambda} = \lambda + \rho \mu + O(\rho^2)$. We may assume $\|v\| = \|\tilde{v}\| = 1$, so $v^T w = 0$. If $\lambda$ is a simple eigenvalue of $B$, $B - \lambda I$ is invertible on the orthogonal complement $v^\perp$ of $v$. Taking $(B + \rho b b^T) \tilde{v} = \tilde{\lambda} \tilde{v}$ to first order in $\rho$, we have $$B w + b b^T v = \mu v + \lambda w$$ Let $c = b - (b^T v) v$ which is orthogonal to $v$, so that this equation splits into $$B w + (b^T v) c = \lambda w \ \text{and} (b^T v)^2 v = \mu v$$ Thus $\mu = (b^T v)^2$ and $w = - (b^T v) (B - \lambda I)|_{v^\perp}^{-1} c$.