Matrix Multiplication is not Commutative/Order 2 Square Matrices

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Theorem

Let $R$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.

Let $\map {\MM_R} 2$ denote the $2 \times 2$ matrix space over $R$.

The operation of (conventional) matrix multiplication is not commutative over $\map {\MM_R} 2$.


Proof

As $R$ is a ring with unity, we have that:

\(\ds 0_R\) \(\ne\) \(\ds 1_R\)
\(\ds 0_R \times 0_R\) \(=\) \(\ds 0_R\)
\(\ds 0_R \times 1_R\) \(=\) \(\ds 0_R = 1_R \times 0_R\)
\(\ds 1_R \times 1_R\) \(=\) \(\ds 1_R\)


Now let:

\(\ds \mathbf A\) \(=\) \(\ds \begin {pmatrix} 0_R & 1_R \\ 0_R & 0_R \end {pmatrix}\)
\(\ds \mathbf B\) \(=\) \(\ds \begin {pmatrix} 0_R & 0_R \\ 1_R & 0_R \end {pmatrix}\)


By definition, both $\mathbf A$ and $\mathbf B$ are elements of $\map {\MM_R} 2$.

It will be demonstrated that $\mathbf A$ and $\mathbf B$ do not commute.


We have:

\(\ds \mathbf A \mathbf B\) \(=\) \(\ds \begin {pmatrix} 0_R & 1_R \\ 0_R & 0_R \end {pmatrix} \begin {pmatrix} 0_R & 0_R \\ 1_R & 0_R \end {pmatrix}\)
\(\ds \) \(=\) \(\ds \begin {pmatrix} 0_R \times 0_R + 1_R \times 1_R & 0_R \times 0_R + 1_R \times 0_R \\ 0_R \times 0_R + 0_R \times 1_R & 0_R \times 0_R + 0_R \times 0_R \end{pmatrix}\)
\(\ds \) \(=\) \(\ds \begin {pmatrix} 1_R & 0_R \\ 0_R & 0_R \end{pmatrix}\)


and:

\(\ds \mathbf B \mathbf A\) \(=\) \(\ds \begin {pmatrix} 0_R & 0_R \\ 1_R & 0_R \end {pmatrix} \begin {pmatrix} 0_R & 1_R \\ 0_R & 0_R \end {pmatrix}\)
\(\ds \) \(=\) \(\ds \begin {pmatrix} 0_R \times 0_R + 0_R \times 0_R & 0_R \times 1_R + 0_R \times 0_R \\ 1_R \times 0_R + 0_R \times 0_R & 1_R \times 1_R + 0_R \times 0_R \end{pmatrix}\)
\(\ds \) \(=\) \(\ds \begin {pmatrix} 0_R & 0_R \\ 0_R & 1_R \end{pmatrix}\)


and it is seen that:

$\mathbf A \mathbf B \ne \mathbf B \mathbf A$


Thus, whatever the nature of the ring with unity $R$, it is never the case that matrix multiplication is commutative over $\map {\MM_R} 2$.

$\blacksquare$


Sources

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