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Suppose $A$ is a square matrix over complex numbers and $u$ is an eigenvector with eigenvalue $\alpha$. Consider perturbing $A$ using $u$ to get $B = A + uv^H$ for some vector $v$. Then we would like to relate the characteristic polynomial of $A$ with that of $B$ as

$(\lambda - \alpha)$ $det(\lambda I_n - B)$ = $(\lambda - \alpha - v^Hu)$ $det(\lambda I_n - A)$

Any hints will be appreciated.

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Hint: Let $P$ the base change matrix such that $J=PAP^*$ is in Jordan canonical form and $Pu=e_1$ (check that you can make that choice!). Further set $v'=P^{-H}v$. Then we have $PBP^{*}=J+e_1v'^H$. Since the determinant is multiplicative we can assume wlog that $A=J$ and $B=J+e_1v^H$.

Now all matrices involved are very simple and you should be able to read off the determinants directly. Can you finish from here?

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  • $\begingroup$ Got it! Thanks for the hint. $\endgroup$ – BharatRam Sep 4 '12 at 4:52

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