# Explain eigenvalues of a distance/cost matrix

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

• Is the cost calling from $i$ to $j$ is equal to the cost calling from $j$ to $i$?
– Lee
May 21, 2020 at 17:51
• @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 May 21, 2020 at 17:54
• 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
– Lee
May 21, 2020 at 17:59
• The largest eigenvalue correspond to a nonnegative eigenvector which likely has some nice interpretation (you can have a look at eigencentrality)
– Surb
Jun 5, 2020 at 9:14
• You may also be interested in looking at the eigenvalues of the graph Laplacian.
– Surb
Jun 5, 2020 at 9:16

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}