Comment author:ESRogs
10 October 2017 03:08:46PM
0 points
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I would expect Tim to have reliably ended up with a negative score on his FIRST eigendemocracy vector, who would also probably be unusually high (maybe the highest user) on a second or third such vector.

Is there a natural interpretation of what the first vector means vs what the second or third mean? My lin alg is rusty.

Assuming the interaction matrix is diagonizable, the system state can be represented as a linear combination of the eigenvectors. The eigenvector with the largest positive eigenvalue grows the fastest under the system dynamics. Therefore, the respective compontent of the system state will become the dominating component, much larger than the others. (The growth of the components is exponential.) Ultimately, the normalized system state will be approximately equal to the fastest growing eigenvector, unless there are equally strongly growing other eigenvectors.

If we assume the eigenvalues are non-degenerate and thus sortable by size, one can identify the strongest growing eigenvector, the second strongest growing eigenvector, etc. I think this is what JenniferRM means with 'first' and 'second' eigenvector.

## Comments (294)

BestIs there a natural interpretation of what the first vector means vs what the second or third mean? My lin alg is rusty.

I wondered the same thing. The explanation I've come up with is the following:

See https://en.wikipedia.org/wiki/Linear_dynamical_system for the relevant math.

Assuming the interaction matrix is diagonizable, the system state can be represented as a linear combination of the eigenvectors. The eigenvector with the largest positive eigenvalue grows the fastest under the system dynamics. Therefore, the respective compontent of the system state will become the dominating component, much larger than the others. (The growth of the components is exponential.) Ultimately, the normalized system state will be approximately equal to the fastest growing eigenvector, unless there are equally strongly growing other eigenvectors.

If we assume the eigenvalues are non-degenerate and thus sortable by size, one can identify the strongest growing eigenvector, the second strongest growing eigenvector, etc. I think this is what JenniferRM means with 'first' and 'second' eigenvector.