# Spectral measure of a finite graph

Let $$A$$ be the adjacency operator of connected, locally finite graph $$G = (V,E)$$ ($$A$$ seen as an operator on $$\ell^2(V)$$). Then we have the spectral representation $$A = \int_{\sigma(A)} t \mu(dt)$$ where $$\mu$$ is a resolution of the identity. For $$u, v \in V$$, let $$e_u$$ be the vector in $$\ell^2(V)$$ whose $$u$$-entry is $$1$$ and all other entries are $$0$$. Define the spectral measures by $$\mu_{u,v}(dt) = \langle \mu(dt) e_u, e_v \rangle.$$ In this case, $$\mu_{u,u}$$ is the unique probability measure on $$\mathbb{R}$$ such that for all integers $$k \geq 1$$, $$\int_{\sigma(A)} t^k \mu_{u,u}(dt) = \langle A^k e_u, e_u \rangle.$$

Question: if $$|V|$$ is finite, then $$A$$ is a symmetric matrix (and self-adjoint), and the spectrum is discrete. If $$(v_1, \ldots, v_n)$$ is an orthonormal basis of eigenvectors associated to the eigenvalues $$(\lambda_1, \ldots, \lambda_n)$$, how can one show from the definition that $$\mu_{u,u} = \sum_{k=1}^n \langle v_k, e_u \rangle^2 \delta_{\lambda_k},$$ or I think equivalently that $$\mu = \frac{1}{n} \sum_{i=1}^n \delta_{\lambda_i},$$ especially if one should be able to recover $$A$$ from $$\mu$$?