Here is a question from a banach algebra course,that i got stuck into:
Consider the Banach algebra $A=\{T_a ,a\in \mathbb{C^n}\}$ where $T_a:\mathbb{C^{n+1}}\longrightarrow \mathbb{C^{n+1}}$ is the $a$-weighted right shift operator ,namely $$ T_a(c_0,...,c_n)=(0,a_1c_1,...,a_nc_n)$$ The point is to identify the characters of this algebra ,that is every linear and multiplicative functional from $A$ to $\mathbb{C}$.
Clearly every $f_k(T_a)=a_k$ defines one character .On the other hand i would like to show that every other character is of this form. So considering $f:A\longrightarrow \mathbb{C}$ to be a character it has to obey $$ (*) f(T_aT_a)=f^2(T_a)$$It must be enough to calculate over the standard basis of $\mathbb{C^n}$ .So since $T_{e_k}(e_k)=e_{k+1}$ $(*)$ gives $$f(T_{e_k}T_{e_k}(e_k))=f(T_{e_k}(e_{k+1}))=0=f^2(e_{k+1})$$. Does this lead to a coclusion or am i totally wrong here? Any hint would be great...
I don't really see why you call your operator a "shift". Your $Ta$ is implemented by the diagonal matrix with diagonal $(0,a_1,\ldots,a_n)$. So the algebra will be a subalgebra of $\mathbb C^n$ (it could be smaller if there are repetitions within $a_1,\ldots,a_n$).
By seeing the algebra as $\mathbb C^d$, it is easy to check that the characters are precisely the projections onto the components. As you say, they are induced by $T_a\longmapsto a_k$, $k=1,\ldots,n$.