Eigendecomposition of sum of rank 1 matrices

Question

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 $yy^T$?

Answer 1

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 $$ \operatorname{tr}(C^2) = \lambda_1^2 + \lambda_2^2 =\|x\|^4 + \|y\|^4 + 2 (x^T y)^2, $$ by the quadratic formula, we have for $i=1,2$, $$ \lambda_i = \frac{\|x\|^2+\|y\|^2}{2} \pm \frac{1}{2}\sqrt{(\|x\|^2-\|y\|^2)^2+4(x^T y)^2}, $$ 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 $\|v_i\|^2=1$.