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Is it possible to find an analytical expression for the innverse of the matrix $A$ defined by :

$$i\neq j \;:\; A_{ij}=x_ix_j$$ $$A_{ii}=\alpha x_i$$

With $\alpha$ a constant and $x$ a vector. If yes, how could I do that please ?

J.A
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  • Why do you think it is invertible? It looks like an outer product matrix, which typically has rank 1. –  Jun 03 '20 at 22:21
  • I assume it's invertible since it is a symmetrical matrix – J.A Jun 03 '20 at 23:32
  • A matrix of all ones is symmetrical but not invertible. Positive semi-definite matrices, in general, might not be invertible, since the eigenspace associated with the zero eigenvalue might not have sufficient rank. –  Jun 03 '20 at 23:38

1 Answers1

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Your matrix is $D+xx^T$ where $D$ is the diagonal matrix such that $d_{ii}=\alpha x_i-x_i^2$. This is a rank-$1$ update of $D$. Provided that $D$ is nonsingular, $D+xx^T$ is invertible if and only if $1+x^TD^{-1}x\ne0$. If this is the case, you may apply Sherman-Morrison formula to obtain $$ A^{-1}=(D+xx^T)^{-1}=D^{-1}-\frac{D^{-1}xx^TD^{-1}}{1+x^TD^{-1}x}. $$

user1551
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