Let $A$ be a self-adjoint operator on .the Hilbertspace $H$ and let $\lambda_0$ be an eigenvalue of $A$ with corresponding eigenvector $\psi$. The spectral theorem tells us,that there is a projection-valued measure $P_A$ such that $A = \int \lambda dP_A(\lambda)$. To each vector $x \in H$ we associate a finite Borel measure $\mu_x$ defined by $\mu_\psi(\Omega) = \langle \psi, P_A(\Omega) \psi\rangle$.
1) I would like first to understand the very basic example of $A$ an operator on finite-dimensional $\mathbb C^n$ with eigenvalues $\lambda_1,...,\lambda_m$, $P_j$ the projection corresponding to eigenspace of $\lambda_j$. Then $A = \sum_j \lambda_j P_j$. I would like to understand why then we have $\mu_\psi(\lambda) = \sum \|P_j \psi \|^2 \delta(\lambda-\lambda_j)$.
2) Secondly, for a self-adjoint operator $A$ on $H$, if $\psi$ is an eigenvector to eigenvalue $\lambda_0$, $\|\psi\| = 1$, is it possible to give $\mu_\psi$ explicitely? Maybe someone can give a short explanation what I have to understand to find the expression by myself.
Edit: I feel stupid, but one last question: why is $\mu_\psi(\{\lambda_0\}) := \langle \psi, P_A(\lambda_0\}) \psi \rangle = \|\psi\|^2$? (in the case 2 that $\psi$ is an eigenvector to the eigenvalue $\lambda_0$)