I have a question regarding the Woodbury matrix identity. In general, the Woodbury matrix identity is as follows

$(\mathbf{A} + \mathbf{u} \mathbf{u}^T)^{-1} = \mathbf{A}^{-1} - \frac{\mathbf{A}^{-1}\mathbf{u} \mathbf{u}^T\mathbf{A}^{-1}}{1 + \mathbf{u}^T \mathbf{A}^{-1}\mathbf{u} }$ where $\mathbf{u}$ is a $n \times 1$ vector and $\mathbf{A}$ is a $n \times n$ matrix.

Is there any formula to calculate the sum of this expression? i.e., $(\mathbf{I} + \sum_{k = 1}^{K}a_k \mathbf{u}_k \mathbf{u}_k^T)^{-1}$ where $\mathbf{I}$ is an identity matrix of size $n \times n$?

Thank you very much in advance


1 Answer 1


The more general Woodbury matrix identity states that $$\left(A + UCV \right)^{-1} = A^{-1} - A^{-1}U \left(C^{-1} + VA^{-1}U \right)^{-1} VA^{-1}$$ where $A$ is an $n \times n$ matrix, $C$ is a $k \times k$ matrix, $U$ is an $n \times k$ matrix, and $V$ is a $k \times n$ matrix.

If we set $A = I$, $C = \text{diag}(a_1,\ldots,a_k)$, $U = \begin{bmatrix}u_1 & \cdots & u_k\end{bmatrix}$, and $V = U^T$, we obtain the formula:

$$\left(I + UCU^T \right)^{-1} = I - U \left(C^{-1} + U^TU \right)^{-1} U^T,$$ where the left side is equal to the expression you are interested in, and the right side only requires inverting a $k \times k$ matrix instead of an $n \times n$ matrix.


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