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I am trying to find the inverse of the following matrix \begin{equation} \mathbf{X} = \sum_{r=1}^R \xi_{r} \mathbf{a}_{N, r} \mathbf{a}_{N, r}^H + \chi \mathbf{I}_N \end{equation} 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

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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 $$ X^{-1} = \chi^{-1} I_N - \chi^{-2} A(\Xi^{-1} + \chi^{-1}A^HA)^{-1}A^H \\ = \chi^{-1} \left[I_N - A(\chi\cdot \Xi^{-1} + A^HA)^{-1}A^H\right] $$

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