I'm having a little trouble finding the eigenvectors and eigenvalues of this covariance matrix I have: $$ \mathbf{B}\mathbf{B}'a + \mathbf{R}. $$ $\mathbf{R}$ is diagonal, $\mathbf{B}$ is some low-rank matrix, and $a$ is a positive real number.

This is as far as I got. In the case that $\mathbf{B}$ is a column vector, we can use the Matrix Determinant Lemma to simplify the characteristic polynomial: \begin{align*} 0 &= \det\left(a \mathbf{B}\mathbf{B}' + \mathbf{R} - \lambda I \right)\\ &= \prod_{i=1}^n(R_{ii} - \lambda)\left( 1 + a\sum_{i=1}^n \frac{B_i^2}{R_{ii}-\lambda} \right)\\ &= \prod_{i=1}^n(R_{ii} - \lambda)\left(\frac{ \prod_{i=1}^n(R_{ii}-\lambda) + a\sum_{i=1}^nB_i^2\prod_{j\neq i}(R_{jj} - \lambda) }{ \prod_{i=1}^n(R_{ii}-\lambda) } \right) \\ &= \prod_{i=1}^n(R_{ii}-\lambda) + a\sum_{i=1}^nB_i^2\prod_{j\neq i}(R_{jj} - \lambda). \end{align*}

I was thinking a lot of the eigenvalues would be something close to the diagonal elements of $\mathbf{R}$ because $$ \left[\mathbf{B}\mathbf{B}'a + \mathbf{R}\right]\left[\mathbf{I} - \frac{\mathbf{B}\mathbf{B}'}{\mathbf{B}'\mathbf{B}}\right] = \mathbf{R}\left[\mathbf{I} - \frac{\mathbf{B}\mathbf{B}'}{\mathbf{B}'\mathbf{B}}\right] , $$ but $\mathbf{R}\left[\mathbf{I} - \frac{\mathbf{B}\mathbf{B}'}{\mathbf{B}'\mathbf{B}}\right] \neq \left[\mathbf{I} - \frac{\mathbf{B}\mathbf{B}'}{\mathbf{B}'\mathbf{B}}\right]\mathbf{R}$ and $\mathbf{I} - \frac{\mathbf{B}\mathbf{B}'}{\mathbf{B}'\mathbf{B}}$ isn't even full rank.

  • 2
    $\begingroup$ It's unclear what would constitute "finding" the eigensystem, especially since you're vague about what $B$ is. You cannot reasonably hope for an effective formula when the question is stated in such generality. Perhaps something like Weyl's Inequalities will help you establish useful bounds. $\endgroup$ – whuber Feb 7 '18 at 22:49

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