TITLE: Eigenvalues of a sum of rank-one matrices?
QUESTION [10 upvotes]: 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.

REPLY [16 votes]: 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 $0$.