Unital and commutative of a subset of a Banach algebra

Question

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.

Answer 1

What appears to be is what is: the identity matrix $I \in M_3(\Bbb C)$ is indeed the unit for $\mathcal A$; see item (5) below.

With

$x_0 = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \in \mathcal A, \tag 1$

it follows from a very easy direct calculation that

$x_0^2 = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}; \tag 2$

thus

$\mathcal A \ni A = \begin{bmatrix} \alpha & \beta & \gamma \\ 0 & \alpha & \beta \\ 0 & 0 & \alpha \end{bmatrix}$ $= \alpha \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} + \beta \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} + \gamma \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = \alpha I + \beta x_0 + \gamma x_0^2; \tag 3$

thus we see that

$\mathcal A = \text{span}(\{I, x_0, x_0^2\}). \tag 4$

$I \in \mathcal A$ is a unit for $\mathcal A$, since

$A \in \mathcal A \Longrightarrow AI = IA = A, \tag 5$

a property clearly inherited from $M_3(\Bbb C)$.

It is probably worth showing that $\mathcal A$ is multiplicatively closed; we begin with the observation that

$x_0^3 = x_0 x_0^2 = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = 0, \tag 6$

from which it follows that

$x_0^k = x_0^{k - 3}x^3 = 0, \; k \ge 3; \tag 7$

thus we have for $\alpha, \beta, \gamma \in \Bbb C$,

$x_0(\alpha I + \beta x_0 + \gamma x_0^2) = \alpha x_0 + \beta x_0^2 \in \mathcal A, \tag 8$

$x_0^2(\alpha I + \beta x_0 + \gamma x_0^2) = \alpha x_0^2 \in \mathcal A; \tag 9$

it is easy to see that (8) and (9) in concert show that any product of elements of $\mathcal A$,

$(\alpha I + \beta x_0 + \gamma x_0^2)(\rho I + \sigma x_0 + \tau x_0^2) \in \mathcal A. \tag{10}$

Since every element of $\mathcal A$ is of the form $\alpha I + \beta x_0 + \gamma x_0^2$, and all powers of $x_0$ (including $x_0^0 = I$) commute, it follows that $\mathcal A$ is commutative.

Now if I recall correctly the characters of $\mathcal A$ are the multiplicative linear functionals $\theta: \mathcal A \to \Bbb C$; that is, linear functionals such that also

$\theta(uv) = \theta(u) \theta(v) \in \Bbb C; \tag{11}$

i.e., the $\theta$ are simply the $\Bbb C$-algebra homomorphisms $\mathcal A \to C$. For any such $\theta$ we have

$(\theta(I))^2 = \theta(I^2) = \theta(I) \in \Bbb C; \tag{12}$

this implies

$\theta(I) \in \{0, 1\} \subset \Bbb C; \tag{13}$

if

$\theta(I) = 0, \tag{14}$

then for any $a \in \mathcal A$ we have

$\theta(a) = \theta(aI) = \theta(a) \theta(I) = 0; \tag{15}$

that is, $\theta$ is the trivial algebra homomorphism $\mathcal A \to \Bbb C$; we thus assume that

$\theta(I) = 1; \tag{16}$

we next note that if $v \in \mathcal A$ is nilpotent,

$v^k = 0, \; k \in \Bbb N, \tag{17}$

then

$(\theta(v))^k = \theta(v^k) = \theta(0) = 0 \Longrightarrow \theta(v) = 0; \tag{18}$

that is, every nilpotent

$v \in \ker \theta; \tag{19}$

we may now compute the $\theta$ via the observation that any

$\alpha I + \beta x_0 + \gamma x_0^2 = \alpha I + v, \tag{20}$

where

$v = \beta x_0 + \gamma x_0^2 \tag{21}$

satisfies

$v^3 = x_0^3(\beta I + \gamma x_0)^3 = 0; \tag{22}$

since $v$ is nilpotent we find

$\theta(\alpha I + v) = \alpha \theta(I) + \theta(v) = \alpha; \tag{23}$

this shows the only non-trivial character of $\mathcal A$ is

$\theta(\alpha I + \beta x_0 + \gamma x_0^2) = \alpha; \tag{24}$

we leave to the reader the simple task of showing that $\theta$ is in fact an algebra homomorphism $\mathcal A \to \Bbb C$.

In closing, we note that $\mathcal A$ is Banach: it is closed, being a finite-dimensional subspace of the finite-dimensional Banach algebra $M_3(\Bbb C)$.

Answer 2

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 $\alpha1+\beta x_0+\gamma x_0^2\mapsto\alpha$.