How to decompose a representation into direct sum of cyclic representation? Let $U$ be the bilateral shift operator on $\ell^2(\mathbb Z)$, and let $T=U+U^*$. How to calculate the spectrum $\sigma(T)$? And how to show there is no cyclic vector for the action of $C^*(T,I)$? Further how to decompose this representation to direct sum of cyclic representations?
Something I know so far: if there is a cyclic vector for the action of $C^*(T,I)$, then the commutant $C^*(T,I)'$ will be commutative. But I don't know how to identify this commutant.
 A: Mapping $\ell^2(\mathbb Z)$ to $L^2(S^1)$ by $(a_k)\mapsto \sum a_k z^k$, $U$ is unitarily equivalent to multiplication by $z$, and $T$ is unitarily equivalent to multiplication by $2\mathrm{Re}(z)$.  The spectrum of a multiplication operator $M_f$ on $L^2(S^1)$ when $f$ is continuous is its range, so the spectrum of $T$ is $[-2,2]$.   The C*-algebra generated by $T$ is unitarily equivalent to the subalgebra of mulitplication operators on $L^2(S^1)$ generated by $I$ and $M_{\mathrm Re(z)}$.  These will have the form $M_f$ with $f(z)=f(\overline{z})$.  The commutant is not commutative because it contains all of $C(S^1)$ acting as multiplication operators, and it contains the map $V\in B(L^2(S_1))$ such that $Vg(z)=g(\overline{z})$.  
You can write $L^2(S^1)$ as a direct sum of the eigenspaces for eigenvalues $1$ and $-1$ of $V$, each invariant for $M_{\mathrm{Re}(z)}$.  Explicitly, for each $g$ in $L^2$, $g=h+k$, $Vh=h$, $Vk=-k$, with $h(z)=\frac12(g(z)+g(\overline z))$ and $k(z)=\frac12(g(z)-g(\overline z))$, i.e., $h=\frac12(I+V)g$ and $k=\frac12(I-V)g$.  The subspace $(I+V)L^2$ has cycle vector $1$, and $(I-V)L^2$ has cyclic vector $z-\overline z$.
