# Approximation of Woodbury matrix identity

I am trying to find the inverse of the following matrix $$$$\mathbf{X} = \sum_{r=1}^R \xi_{r} \mathbf{a}_{N, r} \mathbf{a}_{N, r}^H + \chi \mathbf{I}_N$$$$ where $$\mathbf{a}_{N, r}$$ is a $$N$$-dimensional column vector $$\mathbf{I}_N$$ is an identity matrix of size $$N \times N$$. In particular, if $$R = 1$$, $$X^{-1}$$ could be easily found in a closed form using Woodbury matrix identity as a function of $$\mathbf{a}_{N, r}$$ and $$\mathbf{I}_N$$, however, for $$R>1$$, I cannot find a closed form inverse for this matrix with the similar form of $$\mathbf{X}$$ and hence I am looking for an approximation of it. Could anybody please help me in this situation?

Many thanks

Let $$A$$ denote the matrix whose columns are $$a_{N,1},\dots,a_{N,R}$$ and let $$\Xi$$ denote the diagonal matrix whose diagonal entries are $$\xi_1,\dots,\xi_R$$. We can express the matrix in question as $$X = A\Xi A^H + \chi I_N.$$ With this, we can use the Woodbury matrix identity to find $$X^{-1} = \chi^{-1} I_N - \chi^{-2} A(\Xi^{-1} + \chi^{-1}A^HA)^{-1}A^H \\ = \chi^{-1} \left[I_N - A(\chi\cdot \Xi^{-1} + A^HA)^{-1}A^H\right]$$