Eigendecomposition of sum of rank 1 matrices Suppose I have a rank 2 $n$-by-$n$ matrix
$$
C = x x^T + yy^T
$$
where $x,y \in \mathbb{R}^n$. Are the eigenvalues/eigenvectors of $C$ related in any way to the eigenvalues of $xx^T$ and $yy^T$? 
 A: 1. Eigenvalues 
Let $\lambda_1$ and $\lambda_2$ be the non-zero eigenvalues of $C$.
Since 
$$
\operatorname{tr}(C) = \lambda_1 + \lambda_2 = \|x\|^2 + \|y\|^2
$$
and
$$
\operatorname{tr}(C^2) = \lambda_1^2 + \lambda_2^2 =\|x\|^4 + \|y\|^4 - 2 x^T y,
$$
by the quadratic formula, we have for $i=1,2$,
$$
\lambda_i = \frac{\|x\|^2+\|y\|^2}{2} \pm \frac{1}{2}\sqrt{(\|x\|^2-\|y\|^2)^2+4x^T y},
$$
where $\lambda_1$ and $\lambda_2$ correspond to $+$ and $-$.
2. Eigenvectors 
Suppose that we have eigenvectors of a form as follows:
$$
v_i = \alpha_i x + \beta_i y,
$$
where $\alpha$ and $\beta$ are scalars.
Then by the definition of eigenvector,
$$
C v_i = \lambda_i v_i
$$
which is
$$
(xx^T+yy^T)(\alpha_i x + \beta_i y) = \lambda_i(\alpha_i x + \beta_i y).
$$
Rearranging terms,
$$
LHS = \left(\alpha_i\|x\|^2 + \beta_i (x^Ty)\right) x + \left( \beta_i \|y\|^2 + \alpha_i (x^T y)\right) y, \\
RHS = \alpha_i \lambda_i x + \beta_i \lambda_i y,
$$
and
$$
\alpha_i\|x\|^2 + \beta_i (x^Ty)=\alpha_i \lambda_i, \\
\beta_i \|y\|^2 + \alpha_i (x^T y)=\beta_i \lambda_i,
$$
since the coefficient in front of each $x$ and $y$ should be the same.
Therefore the ratio of $\alpha_i$ and $\beta_i$ is
$$
\alpha_i:\beta_i=x^Ty:(\lambda_i-\|x\|^2) \\
\text{(or $\alpha_i:\beta_i=(\lambda_i-\|y\|^2):x^Ty$ gives the same result)},
$$
and the eigenvectors are
$$
v_i=\mathcal{N}_i \left((x^Ty) x + (\lambda_i-\|x\|^2) y \right), 
$$
where $\mathcal{N}_i$ is the normalization constant to make $\|v_i\|^2=1$.
