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Consider relations between people is defined by a weighted symmetric undirected graph $W$, and $w_{ij}$ shows amount of weight $i$ has for $j$. Assume all weights are non-negative and less than $1$ i.e. $$0\leq w_{ij}<1, \forall{i,j}$$ and symmetric $w_{ij}=w_{ji}$. We say $i$ and $j$ are friends if $w_{ij}>0$.

Define the influence matrix $\Psi=[\text{I}+W]^{-1}$ (Assume it is well-defined). Is it always the case that my influence on myself is greater than my influence on my friends? $$\Psi_{i,i}>\sum_{j\in N(i)}|\Psi_{i,j}|$$

where $N(i)$ is a set of friends of $i$.

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Assuming that $w_{ii}=0$, it's not even necessarily true that $\Psi_{ii} > 0$ $$\begin{bmatrix}1 & a & b\\ a & 1 & c\\ b & c & 1\end{bmatrix}^{-1} =\frac{1}{1-a^2-b^2-c^2+2abc}\begin{bmatrix}1-c^2 & bc-a & ac-b\\ bc-a & 1-b^2 & ab-c\\ ac-b & ab-c & 1-a^2\end{bmatrix}$$ So if $1-a^2-b^2-c^2+2abc<0$, then $\Psi$ has negative elements on the diagonal. For example, $$\begin{bmatrix}1 & 4/5 & 4/5\\ 4/5 & 1 & 1/5\\ 4/5 & 1/5 & 1\end{bmatrix}^{-1} =\frac{1}{8}\begin{bmatrix}-120 & 80 & 80\\ 80 & -45 & -55\\ 80 & -55 & -45\end{bmatrix}.$$

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  • $\begingroup$ Yes you are right, thanks for your answer. $\endgroup$ – user54626 Nov 12 '13 at 3:13
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Look at the inverse of $$\left[\begin{array}{ccc}1 & 0 & .9\\ 0 & 1 & .9\\ 0 & .9 & 1\end{array}\right]$$

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  • $\begingroup$ This matrix is not symmetric, which was stated as a condition in the question. Can you make your point with a symmetric matrix? $\endgroup$ – half-integer fan Nov 11 '13 at 23:54
  • $\begingroup$ Yes, as the above comment mentioned, your matrix is not symmetric, but thanks for your comment anyway. $\endgroup$ – user54626 Nov 11 '13 at 23:56

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