Consider the self adjoint matrix $T$
\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 &0 &2 \end{bmatrix} The question is the following: I want to understand the algorithm of Hahn-Hellinger (Spectral Theorem) so far I tried is that since 1 is eigenvalue of multiplicity of 2 so it does not have cyclic vector, so $(1,0,1)$ and $(0,1,0)$ are cyclic vectors breaking the space into two parts. My simple question how to find the spectral representation $\pi: L^\infty(X,\mu) \mapsto B(L^2(X,\mu))$ and unitary such that $\pi_1: C(X) \mapsto B(\mathcal{H})$ are equivalent where $X=\sigma (T)$, $\mu$ is spectral measure
I'm not entirely sure what you are looking for. Here $\sigma(T)=\{1,2\}$. The C$^*$-algebra generated by $T$ is $$ \left\{\begin{bmatrix} a&0&0\\ 0&a&0\\ 0&0&b\end{bmatrix}:\ a,b\in\mathbb C\right\}\simeq\mathbb C^2, $$ while $$ C(\sigma(T))=C(\{1,2\})\simeq \mathbb C^2. $$ The Gelfand transform here is the map $\Gamma: C^*(T)\to C(\sigma(T))$, which in this case is explicitly given by $$ \Gamma\left(\begin{bmatrix} a&0&0\\ 0&a&0\\ 0&0&b\end{bmatrix}\right)=\begin{bmatrix} a\\ b\end{bmatrix}. $$ The natural embedding of $C(\sigma(T))$ as multiplication operators on $B(L^2(\sigma(T))$ is given by $$ \pi_1(a,b)=\begin{bmatrix} a&0\\0&b\end{bmatrix}\in M_2(\mathbb C)=B(\mathbb C^2). $$