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The Sherman–Morrison formula states that $$\left(\mathbf{A} + \mathbf{u}\mathbf{v}^T\right)^{-1} = \mathbf{A}^{-1} - \frac{\mathbf{A}^{-1}\mathbf{u}\mathbf{v}^T\mathbf{A}^{-1}}{1 + \mathbf{v}^T \mathbf{A}^{-1} \mathbf{u} }.$$ My question is that $\mathbf{A}$, $\mathbf{u}$ and $\mathbf{v}$ have same assumption with Sherman–Morrison formula, and how can I compute or approximate the inverse $$\left(\mathbf{A} + \mathbf{u}\mathbf{v}^T + \lambda \mathbf{I}\right)^{-1},$$ where $\lambda > 0$ is very small constant.

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  • $\begingroup$ Suppose that $$A_t = \alpha A_{t-1} + \beta u_t v_t^T + \lambda I.$$ Like Sherman–Morrison formula, given $A_{t-1}^{-1}$, $u_t$ and $v_t$, I want to compute the inverse of $A_t$ efficiently, $\endgroup$ – ruobot Nov 16 '16 at 11:59
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Given $B$ any invertible matrix with $\mu\ne 0$ minimum eigenvalue, then $1/\mu$ is the spectral radius of $B^{-1}$, so, if $\lambda<\mu$, we have

$$ (B+\lambda I)^{-1} = B^{-1}\sum_{i=0}^{\infty} B^{-i}\lambda^i(-1)^i $$

so you can approximate the inverse with a partial sum of the series, with $B=A+uv^T$.

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  • $\begingroup$ Thanks for your answer, and please see my comment above. Thanks again! $\endgroup$ – ruobot Nov 16 '16 at 12:09

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