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}.

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