Eigen Values of Sum of Negative Definite Matrix and a rank-one Positive SemiDefinite Matrix , i.e. ($A+xx^{H}$) 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 $B$
 A: 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.
A: 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 $\pmatrix{a &b \\ c & d}$
