## Exhibit the idempotents in a given semigroup of matrices

Let $S \subseteq \mathsf{Mat}_2([-1,1])$ denote the set of all $2 \times 2$ matrices with entries in $[-1,1] \subseteq \mathbb{R}$ such that one row consists of all 0s. Show that $S$ is a semigroup under matrix multiplication and exhibit its idempotents.

The elements of $S$ come in two flavors, $A$ and $B$, where $A = \left\{ \begin{bmatrix} a & b \\ 0 & 0 \end{bmatrix} \ |\ a,b \in [-1,1] \right\}$ and $B = \left\{ \begin{bmatrix} 0 & 0 \\ a & b \end{bmatrix} \ |\ a,b \in [-1,1] \right\}$. Clearly $A \cap B = 0$. (If $A$ is vanilla and $B$ is chocolate, then the zero matrix is swirl.)

Since $S$ is a subset of the known semigroup $\mathsf{Mat}_2(\mathbb{R})$ under multiplication, it suffices to show that $S$ is closed. To that end, it suffices to consider four cases. In what follows, let $a_i,b_i \in [-1,1]$. (Note that $[-1,1]$ is closed under multiplication.)

• $\begin{bmatrix} a_1 & b_1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} a_2 & b_2 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} a_1a_2 & a_1b_2 \\ 0 & 0 \end{bmatrix} \in S$
• $\begin{bmatrix} a_1 & b_1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 \\ a_2 & b_2 \end{bmatrix} = \begin{bmatrix} b_1a_2 & b_1b_2 \\ 0 & 0 \end{bmatrix} \in S$
• $\begin{bmatrix} 0 & 0 \\ a_1 & b_1 \end{bmatrix} \begin{bmatrix} a_2 & b_2 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ a_1a_2 & a_1b_2 \end{bmatrix} \in S$
• $\begin{bmatrix} 0 & 0 \\ a_1 & b_1 \end{bmatrix} \begin{bmatrix} 0 & 0 \\ a_2 & b_2 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ b_1a_2 & b_1b_2 \end{bmatrix} \in S$

So $S$ is indeed a semigroup.

Suppose now that $M = \begin{bmatrix} a & b \\ 0 & 0 \end{bmatrix} \in A$ is a nonzero idempotent. Then we have $M = M^2 = \begin{bmatrix} a^2 & ab \\ 0 & 0 \end{bmatrix}$, and comparing entries, $a = a^2$ and $b = ab$. Now $a(a-1) = 0$ in $\mathbb{R}$, so that either $a = 0$ or $a = 1$. But if $a = 0$, then $b = 0$, so $M = 0$. Thus $a = 1$. Indeed, $\begin{bmatrix} 1 & b \\ 0 & 0 \end{bmatrix}$ is idempotent.

Similarly, if $M = \begin{bmatrix} 0 & 0 \\ a & b \end{bmatrix} \in B$ is a nonzero idempotent, then $b = 1$, and conversely.

So the idempotents in $S$ are either 0, $\begin{bmatrix} 1 & b \\ 0 & 0 \end{bmatrix}$ for some $b \in [-1,1]$, or $\begin{bmatrix} 0 & 0 \\ a & 1 \end{bmatrix}$ for some $a \in \mathbb{R}$.