# Eigenvectors for the sum of two symmetric matrices

$R = S + xx^t$

where $x \in \mathbb{R}^n$ and $R$ and $S$ are $n \times n$ covariance matrices.

Is there anything I can say about the eigenvectors of $R$ and $S$? Or at least the largest eigenvalue and eigenvector pair of R?

Is there a way to express the eigenvectors of $R$ in terms of $S$ and $x$?

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The sum of two covariance matrices is positive semidefinite; the eigenvalues are non-negative. –  Emre May 31 '12 at 2:11

The rank one matrix $x x^t$ has a single nonzero eigenvalue $\lambda=|x|^2$, with $x$ itself as eigenvector, because $(x x^t) x = |x|^2 x$.

Now, unless $x$ happens to be an eigenvector of $S$, I dont' think you can say anything about the eigenvectors of $R$.

Regarding the largest eigenvalue of $R$ (I guess you meant eigenvalue instead of eigenvector?), recall that

$$\lambda^{max}_R = max_{|w|=1} ( w^T R w)$$

but $$w^T R w = w^T S w + w^T x x^t w \le \lambda^{max}_S + |x|^2$$

Further, calling $w_S$ the normalized eigenvector associated to $\lambda^{max}_S$, we have

$$max_{|w|=1} ( w^T R w) \ge w_S^T R w_S = w_S^T S w_S + w_S^T x x^t w_S = \lambda^{max}_S + w_S^T x x^t w_S \ge \lambda^{max}_S$$

becase $w^T x x^t w \ge 0$ $\forall w$

Hence, I think that's about the only thing you can assert:

$$\lambda^{max}_S \le \lambda^{max}_R \le \lambda^{max}_S + |x|^2$$

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Thanks for the great response. I understand that you can assert the upper bound because $w^t$ is maximized for $R$ and not necessarily for $S$, but how did you get to the lower bound $$\lambda^{max}_S \le \lambda^{max}_R$$ –  Swiss Army Man May 31 '12 at 3:35
"but how did you get to the lower bound?" From the same equation. I added a explanation. –  leonbloy May 31 '12 at 11:32