Can a subset of closed walks in a graph (e.g. from certain vertices only) be determined from the eigenvalues?

Question

If $A$ is an adjacency matrix and $\lambda_1, \lambda_2,...,\lambda_n$ are the eigenvalues of $A$, then $Tr(A) = \lambda_1+\lambda_2+...+\lambda_n$ sums all closed walks of length 1 in the graph represented by $A$. $Tr(A^2) = \lambda_1^2+\lambda_2^2+...+\lambda_n^2$ sums all closed walks of length 2. And so on.

Is there any way to determine closed walks from a single vertex only? For example, can the value $a_{11}$ (the first element of the diagonal of $A$), and the value $a_{11}^2$ (the first element of the diagonal of $A^2$), and so on, be derived from the eigenvalues of $A$?

Thank you!

Answer 1

The generating function $W_{a,a}(G,t)$ for the number of closed walks starting at $a$ is given by the \[ W_{a,a}(G,t) = t^{-1}\phi(G\setminus a,t^{-1})/\phi(G,t). \] Hence the number of closed walks at $a$ is determined by the spectrum of $G$ and the spectrum of $G\setminus a$.

For details, you might look in Chapter 4 of my book "Algebraic Combinatorics". (It's not hard to prove, you basically working out an expression for the $(a,a)$-entry of $(tI-A)^{-1}$.)