Eigenvalues of a sum of rank-one matrices? How can I find the eigenvalues of the sum of rank-one matrices $vv^T + ww^T$?
I know that the eigenvalues of each term is $v^Tv$ and $w^Tw$, respectively.
 A: You cannot expect to be able to express the eigenvalues of $vv^T + ww^T$ from the eigenvalues of each of the two summands. 
For instance, given  $t\in[0,1]$ let
$$
v=\begin{bmatrix}1\\ 0\end{bmatrix}, \ \ w=\begin{bmatrix}t\\ \sqrt{1-t^2}\end{bmatrix}.
$$
Then $vv^T$, $ww^T$ are rank-one projections, each with eigenvalues $\{0,1\} $ which "forget" about $t$. But
$$
vv^T+ww^T=\begin{bmatrix}1+t^2&t\sqrt{1-t^2}\\ t\sqrt{1-t^2}&1-t^2\end{bmatrix},
$$
which has eigenvalues $1+t,1-t$. 
A: Let $A=vv^T+ww^T$, then $A$ has at most rank $2$. So we want to know the potentially non-zero eigenvalues.
We have $\operatorname{Tr}(A)=|v|^2+|w|^2$ and $\operatorname{Tr}(A^2)=|v|^4+|w|^4+2\langle v,w\rangle^2$, so these eigenvalues must satisfy $\lambda_1+\lambda_2=|v|^2+|w|^2$ and $\lambda_1\lambda_2=|v|^2|w|^2-\langle v,w\rangle^2$.
So $\lambda_1$ and $\lambda_2$ are the roots of $X^2-(|v|^2+|w|^2)X+|v|^2|w|^2-\langle v,w\rangle^2$.
The discriminant is $$\Delta=(|v|^2+|w|^2)^2+4\langle v,w\rangle^2-4|v|^2|w|^2=(|v|^2-|w|^2)^2+4\langle v,w\rangle^2>0$$
(or equal to $0$ if $|v|=|w|$ and $v$ and $w$ are orthogonal; in this case we have only one root).
So
$$\lambda_1=\frac{|v|^2+|w|^2-\sqrt{(|v|^2-|w|^2)^2+4\langle v,w\rangle^2}}2$$
and
$$\lambda_2=\frac{|v|^2+|w|^2+\sqrt{(|v|^2-|w|^2)^2+4\langle v,w\rangle^2}}2,$$
the other eigenvalues being $0$.
A: Let $u_1$ and $u_2$ be orthonormal vectors such that $v = a u_1$ and $w = b u_1 + c u_2$ for some scalars $a,b,c$.  Thus in an orthonormal basis starting with $u_1, u_2$, the matrix of 
$v v^T + w w^T$ has first two rows and columns $\pmatrix{a^2 + b^2 & bc \cr bc & c^2\cr}$ and everything else $0$.  The nonzero eigenvalues of your matrix are the eigenvalues of this.
