graph theory - Why are Dynkin diagrams characterized by their eigenvalues?

Yes. For example, with quiver representations, we have a formula

$chi(M,N)=dim Hom(M,N)-dim Ext^1(M,N) = sum d_i(M)d_i(N) - sum_{i to j} d_i(M)d_j(N).$

where $d_i(M)$ is the dimension of M at node i.
The proof is to check that it's true for simples, and then note that the category of representations of the path algebra of a quiver has global dimension 1.

So, what you've noted above is that this is positive definite if and only if the graph is Dynkin. Well, what's good about being positive definite? For one thing, if an object has trivial Ext^1 with itself, then it is rigid, it has no deformations. On the other hand, it also must have $chi(M,M)>0$, since Hom always has positive dimension, and $Ext_1(M,M)=0$.

Thus, if our quiver is not Dynkin, it has dimension vectors where no module can be rigid. On the other hand, if you work a bit harder, you can show Gabriel's theorem:

if the graph is Dynkin, every dimension vector has a unique rigid module and this is indecomposible if and only if $chi(M,M)=1$, that is if $M$ is a positive root of the root system.

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