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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 $\mu$?

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