Unital and commutative of a subset of a Banach algebra Consider the subset of the Banach algebra $M_3(\mathbb{C})$
$$
\mathcal{A}=\left \{ \begin{pmatrix}
\alpha &\beta  &\gamma \\ 
0 & \alpha &\beta \\ 
0 & 0 & \alpha
\end{pmatrix}:\alpha,\beta,\gamma\in\mathbb{C} \right \}
$$
and consider $x_0=\begin{pmatrix}
0 &1  &0 \\ 
0 & 0 &1 \\ 
0 & 0 & 0
\end{pmatrix}\in\mathcal{A}$.
i) Show that $\mathcal{A}=\operatorname{span}\{1,x_0,x_0^2\}$, that $\mathcal{A}$ is a unital Banach algebra, and decide if $\mathcal{A}$ is commutative.
ii) Find all characters on $\mathcal{A}$
I just started to study this subject, so I might be a slow-learner. 1) The first part is obvious. I do not really know how to show if $\mathcal{A}$ is unital and commutative. It appears that the identity matrix is the unit.
 A: After noting that
$$
x_0^2=\begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}
$$
and that $x_0^3=0$, you have that
$$
\begin{pmatrix}
\alpha & \beta & \gamma \\
0 & \alpha & \beta \\
0 & 0 & \alpha
\end{pmatrix}
=\alpha 1+\beta x_0+\gamma x_0^2
$$
and, conversely, $\alpha1+\beta x_0+\gamma x_0^2\in\mathcal{A}$, for every $\alpha,\beta,\gamma\in\mathbb{C}$. So $\mathcal{A}=\operatorname{span}\{1,x_0,x_0^2\}$
Closure under difference is obvious, being $\mathcal{A}$ a subspace; also $1\in\mathcal{A}$. As far as products are concerned, you just need to show that products of any two members of $\{1,x_0,x_0^2\}$ belongs to $\mathcal{A}$, which is true because of $x_0^3=0$. Commutativity is also clear, because the elements of the spanning set commute with each other.
Thus $\mathcal{A}$ is a subalgebra of $M_3(\mathbb{C})$ and is closed, being finite dimensional (or by a direct argument with sequences), so it is complete under the induced norm.
Note also that $\{1,x_0,x_0^2\}$ is a basis for $\mathcal{A}$. Thus a linear functional $\chi\colon\mathcal{A}\to\mathbb{C}$ is determined as soon as we assign $\chi(1)=1$, $\chi(x_0)=z$ and $\chi(x_0^2)=z_1$. The assignment $\chi(1)=1$ is needed if $\chi$ has to be a character; on the other hand a character has to satisfy $\chi(x_0^2)=\chi(x_0)\chi(x_0)=z^2$.
Since $x_0^3=0$ we also have
$$
0=\chi(0)=\chi(x_0^3)=\chi(x_0)^3=z^3
$$
so $z=0$. The algebra has a single character $\alpha1+\beta x_0+\gamma x_0^2\mapsto\alpha$.
