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
- 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): commute
- 2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): commute