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Assume there are N countries. The cost of making a phone call from country $i$ to country $j$ is $C_{ij}$. We know that all costs are non-negative.

(Q1) Can you think of a verbal interpretation of eigenvalues of the matrix $C_{ij}$?

(Q2) Does anything change, if we allow weights to be negative?

I am aware that an eigendecomposition of a transformation $T$ is given by $T = R^{-1}DR$, which means that, if a matrix were to be used as a transformation, it could be interpreted as rotation, scaling, and rotation back to the original basis. However, I'm not necessarily using my matrix to transform anything, so my intuition does not quite help

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    $\begingroup$ Is the cost calling from $i$ to $j$ is equal to the cost calling from $j$ to $i$? $\endgroup$ – Lee May 21 at 17:51
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    $\begingroup$ @Lee if it helps you to make progress, you may assume it. Ultimately I'd love to have some intuition for both symmetric and non-symmetric matrices, but if you provide an intuition that works only for symmetric matrices I would also be happy $\endgroup$ – Aleksejs Fomins May 21 at 17:54
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    $\begingroup$ At least we know that sum of eigenvalues are determined by the sum of domestic call cost. And also the product of eigenvalues by the determinant, which I expect to be negative since the international calls are more expensive. Thus if we take only two countries, one eigenvalue must be positive and one negative $\endgroup$ – Lee May 21 at 17:59
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    $\begingroup$ The largest eigenvalue correspond to a nonnegative eigenvector which likely has some nice interpretation (you can have a look at eigencentrality) $\endgroup$ – Surb Jun 5 at 9:14
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    $\begingroup$ You may also be interested in looking at the eigenvalues of the graph Laplacian. $\endgroup$ – Surb Jun 5 at 9:16
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You may want to take a look at eigenvector centrality. This states that the centrality of node $i$ is the weighted average of the centrality of nodes it is connected to: \begin{align} v_i &= \frac{1}{\beta}\sum_{j\in N}C_{ij}v_j\\ \implies \beta\vec{v} &=C\vec{v} \end{align}

Thus the eigenvalue can be interpreted as the *inverse of the weights/frequencies that each node attaches to its neighbor's centralities. Mathematically, though, the eigenvalues determines the spectrum of the graph. Naively speaking, larger eigenvalues should correspond to more compact graphs (e.g. stars).

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  • $\begingroup$ Thanks for your answer. I can see how this line of thought could lead to the answer I want. But at the moment it is too brief. Eigencentrality was already mentioned in the comments. From an answer I would expect a slightly more detailed explanation of eigencentrality and how it applies to this particular problem, as well as a demonstration and/or proof of the conclusions you have made $\endgroup$ – Aleksejs Fomins Jun 12 at 8:06
  • $\begingroup$ This was the only answer posted in the duration of the bounty - that is the main reason why I awarded the bounty here. But if somebody posts something more about this topic, I am certainly wiling to add another bounty and reward a new answer. $\endgroup$ – Martin Sleziak Jun 12 at 9:20

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