Algebraic Structure

2011 midterm exam in Algebraic Structures

1. If \(S = \left\{ \begin{bmatrix} a & 0 & b \\ 0 & c & d \\ 0 & 0 & a \end{bmatrix} \;\middle|\; a,b,c,d \in \mathbb{R} \right\}\), show that S is a subring of \(M_3(\mathbb{R})\), a 3 by 3 matrix over \(\mathbb{R}\).

Solution: Clearly, \(S \subseteq M_3(\mathbb{R})\). We will show that S is also a ring. To test for a subring, we will show that:

a. S is closed under +, \(\bullet\)

b. The additive and multiplicative identity of \(M_3(\mathbb{R})\) is also in S.

c. \(\forall s \in S\), it is true that the additive inverse -s is also in S.

Let \(s_1, s_2 \in S\) such that \(s_1 = \begin{bmatrix} a_1 & 0 & b_1 \\ 0 & c_1 & d_1 \\ 0 & 0 & a_1 \end{bmatrix}\) and \(s_2 = \begin{bmatrix} a_2 & 0 & b_2 \\ 0 & c_2 & d_2 \\ 0 & 0 & a_2 \end{bmatrix}\).

a. Is S closed + ? Yes, because \(s_1 + s_2 = \begin{bmatrix} a_1 + a_2 & 0 & b_1 + b_2 \\ 0 & c_1 + c_2 & d_1 + d_2 \\ 0 & 0 & a_1 + a_2 \end{bmatrix} \in S\).

Is S closed under \(\bullet\) ? Yes, because \(s_1 \bullet s_2 = \begin{bmatrix} a_1a_2 & 0 & b_1b_2 \\ 0 & c_1c_2 & d_1d_2 \\ 0 & 0 & a_1a_2 \end{bmatrix} \in S\).

b. Clearly, \(\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \in S\) and \(\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \in S\).

c. The inverse of \(s_1\) is \(-s_1 = \begin{bmatrix} -a_1 & 0 & -b_1 \\ 0 & -c_1 & -d_1 \\ 0 & 0 & -a_1 \end{bmatrix} \in S\).

\(\therefore\) S is a subring of \(M_3(\mathbb{R})\).