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The problem is the following:

For $i \in [N]$, let $v_i$ be a $1 \times d$ vector and $b_i$ a scalar. Moreover, let $A$ be a $N \times N$ matrix, whose (i, j)-entry is:

$$ a_{ij} = \begin{cases} \langle v_i, v_j \rangle + b_i &\text{if } i = j \\ \langle v_i, v_j \rangle &\text{if } i \neq j \end{cases}$$

Find $A^{-1}$.

For d = 1, we can find a solution using the Sherman-Morisson equation or using the Leibniz formula and properties of the symmetric group. Which give us:

$$ \begin{equation} \begin{split} a_{ij}^{-1} & = \frac{1}{\prod_{k=1}^n b_k + \sum_{k=1}^n v_k^2 \prod_{l=1, l\neq k}^n b_k} \begin{cases} \prod_{k=1, k\neq i}^n b_k + \sum_{k=1, k\neq i}^n v_k^2 \prod_{l=1, l\neq k, l\neq i}^n b_k & \text{if } i = j\\ -v_jv_i \prod_{k=1, k \neq i, k\neq j}^n b_k& \text{otherwise} \end{cases} \\ \end{split} \end{equation}$$

I didn't add the proof to make a smaller post. So now I am trying to derive a general formula for d but I am not really familiar with this field so I am struggling. And to my understanding with $d>1$, the Serman-Morisson equation is not usable anymore. I am not really sure in which direction to look or if any formula can be derived. Any help would be really appreciated :)

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  • $\begingroup$ I think $A$ is not always invertible. $\endgroup$
    – fusheng
    May 15 at 13:30
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    $\begingroup$ Let $V$ be the matrix whose columns are the $v_k$ vectors and $B\!=\!\operatorname{Diag}(b).\;$ Write $A$ in the form $$ A = B + V^TV $$ to which the Sherman-Morrison-Woodbury formula can be directly applied. $\endgroup$
    – greg
    May 15 at 13:32

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