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Assume all matrices I discuss about are complex $N\times N$ matrices. I have a negative definite hermitian matrix $A$ whose eigen values are $(-c\lambda ,-\lambda,\ldots-\lambda) $ where $c$ and $\lambda$ are positive constants. I also have a rank one positive semi-definite hermitian Matrix $xx^{H}$ (whose eigen values clearly are $(||x||^{2},0,0...0)$. I am familiar with weyl's inequalities. I was wondering if one can find a exact formula for the eigen values of the matrix $B=A+xx^{H}$.

EDIT---

(thanks to users @adamW and @David) I have been able to reduce the problem to a seemingly simple one. There is a diagonal matrix $D$ such that all its entries are zero except for the one in the $(1,1)^{th}$ position, say $D(1,1)=c$ (hence eigen values are $(c,0,\ldots,0)$. I have another rank one positive-semi definite matrix $xx^H$ (again, whose eigen values clearly are $(||x||^{2},0,0...0)$. Even now I can't seem to find a clear solution for this seemingly simple problem. To put everything in perspective, I have the following matrix

\begin{align} B &=D+xx^H \\ D &= \left[ \begin{array}{cccc} c & 0 & \ldots & 0 \\ 0 & 0 & \ldots & 0 \\ \ldots & 0 & 0 & 0 \\ 0 & \dots & 0 & 0 \end{array} \right] \end{align}

what are the eigen values of $B$

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    @adam: I had the same intuition, that there would be no simple solution to this, but sometimes our intuitions go astray...2012-10-14

2 Answers 2

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The two matrices play more symmetric roles in this than is apparent from your formulation. We can state the problem more symmetrically and basis-independently as finding the eigensystem of

$B=xx^H+yy^H\;,$

where your formulation arises if we choose a basis in which $y$ is the first basis vector.

Choose a basis in which the first two basis vectors span $x$ and $y$, and solve the resulting two-dimensional eigenvalue problem exactly.

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    @dineshdileep: Yes, that's what I mean.2012-10-14
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The original question (before the edit) involves a full rank matrix (specifically it is similar to $-\lambda \mathbf{I} + \vec{e_0}(\lambda-c\lambda) \vec{e_0}^T$ or in words the identity with the corner adjusted to be $-c$). Finding a rank one update to that is the same problem as the general rank one update for a full rank matrix. I believe they call it something different than the characteristic formula when talking rank one updates (but from what I read and understood it is a full scale polynomial also, just with poles as well as zeros).

Your after edit version involves a lesser rank, so if indeed that is the problem of interest, it is as easy as the 2nd order polynomial-- it is of rank two, your matrix $B = D +xx^H$.

Try this: if $B = yy^T + xx^T$ then solve for the eigenvectors a combination of the two (since any orthogonal to both $x$ and $y$ are obviously with eigenvalue of zero) $\pmatrix{ax^T + by^T \\ cx^T + dy^T}(yy^T + xx^T) = \pmatrix{\lambda_0 & 0\\0 & \lambda_1}\pmatrix{ax^T + by^T \\ cx^T + dy^T}$

A side note, a rank two matrix may be written as $\pmatrix{x & y}\pmatrix{a & b \\ c & d}\pmatrix{x^T\\ y^T}$ and what is left to diagonalize is the $2 \times 2$ matrix $\pmatrix{a &b \\ c & d}$

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    Ah, I see, the deflation was indeed huge here, with the repeated $\lambda$! That has a large dimensional spectrum that can act nicely!2012-10-15