Point spectrum in Hilbert spaces Let $H$ be a Hilbert space and and $T\in B(H)$ be normal and $\sigma_p(T)$ be the point spectrum of $T$ (i.e the set of all eigenvalues of T) and let $E$ denote the spectral measure. I'm trying to prove the followinga) If $\lambda \in \sigma(T)$ then $E(\{\lambda \})$ is the projection on $\{x\in H: (\lambda I-T)x=0\}.$ In particular, $\lambda \in \sigma_p(T)$ iff $E(\{\lambda\})\ne0.$b)If the span of the eigenvectores of $T$ is dense in $H$ then $$T=\sum_{\lambda \in \sigma_p(T)}\lambda E(\{\lambda\}).$$is convergent in the strong operator topology SOT.I'm stuck on this, Thank you for your help.
Edit: Let $H$ be a Hilbert space and $S$ be a topological space, A spectral measure over $S$ is a function $E$ from the boolean algebra of Borel subsets of $S$ to the soet of othogonla projections in $B(H)$ such that 1)$E(\phi)$=0 and $E(S)=I$2)$E(\sigma \cap \delta)=E(\sigma)E(\delta)$ for all $\sigma ,\delta$ of $S$3)for $(\delta_n)$ of disjoint Borel subsets of $S$ we have $E\left(\bigcup_{n=1}^{\infty}\delta_n\right)=\sum_{n=1}^{\infty}E(\delta_n)$ in the SOT.
 A: I suppose you know the spectral theorem and symbolic calculus, which shall be used, and that you are only taking a specific example and sorting it out.
First, let $\lambda\in\sigma_p(T)$, and call $E_\lambda = E(\{\lambda\})$. We shall show that $E_\lambda H = Ker(\lambda 1 - T) = Ran(\bar{\lambda} 1 - T^*)^{\perp}$.
We first show that $E_\lambda H \subset Ran(\bar{\lambda} 1 - T^*)^{\perp}$. Take any $x,y\in H$, and compute $ \langle (\bar{\lambda}1-T^* ) y \,, \, E_\lambda x \rangle = \langle y \,, \, (\lambda 1 - T ) \,E_\lambda x \rangle $. Symbolic calculus gives us $\int (\lambda - \mu)\chi_{\{\lambda\}}(\mu)\langle y, dE(\mu)x\rangle$. Since the integrating function is identically zero, the inner product is zero for all $x,y\in H$, which proves the first inclusion.
For the reverse inclusion, suppose now $\lambda \ne 0$, and take $x\in ker(\lambda 1 - T)$, which means $\lambda x = Tx$. Symbolic calculus for $\lambda x$ implies
$\lambda x = \int\lambda \,dE(\mu)x = \int \chi_{\{\lambda\}}(\mu) \mu \,dE(\mu)x = E_\lambda T x = \lambda x \Rightarrow x = E_\lambda x \subset E_\lambda H.$
When $\lambda = 0$, one way to do it is to shift the spectrum by considering, for example, $S = T - 1$, and then shifting it back after you have your conclusion. This proves that $E_\lambda H = \ker(\lambda 1 -T)$.
Now, we check that $\lambda\in\sigma_p(T)$ iff $E_\lambda \ne 0$. As we saw, if $\lambda\in\sigma_p(T)$, then $E_\lambda$ projects onto a nonzero subspace, and hence is nonzero. For the converse, notice that $(T-\lambda)E_\lambda x = 0$ (check the symbolic calculus), and hence $E_\lambda H$ is the nonzero kernel of $T-\lambda$.
Let us now check item b. If the span of eigenvectors is dense in $H$, then there is a Hilbertian (orthonormal) basis $\{y_\lambda\}$ such that any $y_\lambda$ is a eigenvector of eigenvalue $\lambda \in \sigma_p(T)$ (notation here is loose, we might have repetition of indices and all, but I think you get it, right?). So any vector $x\in H$ can be written as a sum $x = \sum_\lambda \alpha_\lambda y_\lambda$. Since $T$ is continuous, 
$Tx =
T(\sum_\lambda\alpha_\lambda y_\lambda) =
\sum_\lambda \alpha_\lambda T(y_\lambda) = 
\sum_\lambda\alpha_\lambda \lambda E_\lambda (y_\lambda) = 
\sum_\lambda\lambda E_\lambda (\alpha_\lambda y_\lambda) =
\sum_\lambda\lambda E_\lambda (\sum_\mu \alpha_\mu y_\mu) =
\sum_\lambda\lambda E_\lambda x.$ 
Above, the $\mu$-summation can appear since $E_\lambda$ annihilates indices not equal to $\lambda$. Since this is valid for all $x\in H$, this proves the assertion.
Cheers!
