Matrix Multiplication is not Commutative

== Theorem ==
Let $R$ be a ring with unity.
Let $n \in \Z_{>0}$ be a (strictly) positive integer such that $n \ne 1$.
Let $\map {\MM_R} n$ denote the $n \times n$ matrix space over $R$.
Then (conventional) matrix multiplication over $\map {\MM_R} n$ is not commutative:
:$\exists \mathbf A, \mathbf B \in \map {\MM_R} n: \mathbf {A B} \ne \mathbf {B A}$
If $R$ is specifically not commutative, then the result holds when $n = 1$ as well.
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 2}$, let $\map P n$ be the proposition:
:$\exists \mathbf A, \mathbf B \in \map {\MM_R} n: \mathbf {A B} \ne \mathbf {B A}$
=== Edge Cases ===
==== $n = 1$ ====
Consider the case where $n = 1$.
Then:
and it follows that (conventional) matrix multiplication over $\map {\MM_R} 1$ is commutative $R$ is a commutative ring.
==== $R$ not a Ring with Unity ====
Consider the case where $R$ is not a ring with unity, and is a general ring.
Let $R$ be the trivial ring.
From Matrix Multiplication on Square Matrices over Trivial Ring is Commutative:
:$\forall \mathbf A, \mathbf B \in \map {\MM_R} n: \mathbf {A B} = \mathbf {B A}$
Hence the result does not follow for all rings.
It is not established at this point on exactly which rings (conventional) matrix multiplication $\map {\MM_R} n$ commutes.
However, the existence of just one such ring (the trivial ring) warns us that we cannot apply the main result to all rings.
==== Matrices are not Square ====
We note that $\mathbf A \mathbf B$ is defined when:
:$\mathbf A = \sqbrk a_{m n}$ is an $m \times n$ matrix
:$\mathbf B = \sqbrk b_{n p}$ is an $n \times p$ matrix.
Hence for both $\mathbf A \mathbf B$ and $\mathbf B \mathbf A$ to be defined, it is necessary that:
:$\mathbf A = \sqbrk a_{m n}$ is an $m \times n$ matrix
:$\mathbf B = \sqbrk b_{n p}$ is an $n \times m$ matrix
for some $m, n \in \Z_{>0}$.
But in this situation:
:$\mathbf A \mathbf B$ is an $m \times m$ matrix
while:
:$\mathbf B \mathbf A$ is an $n \times n$ matrix
and so if $\mathbf A$ and $\mathbf B$ are not square matrices, they cannot commute.
=== Basis for the Induction ===
$\map P 2$ is the case:
:$\exists \mathbf A, \mathbf B \in \map {\MM_R} 2: \mathbf {A B} \ne \mathbf {B A}$
This is demonstrated in Matrix Multiplication is not Commutative: Order $2$ Square Matrices.
Thus $\map P 2$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\exists \mathbf A, \mathbf B \in \map {\MM_R} k: \mathbf {A B} \ne \mathbf {B A}$
from which it is to be shown that:
:$\exists \mathbf A, \mathbf B \in \map {\MM_R} {k + 1}: \mathbf {A B} \ne \mathbf {B A}$
=== Induction Step ===
This is the induction step:
From the induction hypothesis, it is assumed that there exist $2$ order $k$ square matrices $\mathbf A$ and $\mathbf B$ such that $\mathbf {A B} \ne \mathbf {B A}$.
For an order $n$ square matrix $\mathbf D$, let $\mathbf {D'}$ be the square matrix of order $n + 1$ defined as:
:$d'_{i j} = \begin {cases} d_{i j} & : i < n + 1 \land j < n + 1 \\ 0 & : i = n + 1 \lor j = n + 1 \end{cases}$
Thus $\mathbf D'$ is just $\mathbf D$ with a zero row and zero column added at the ends.
We have that $\mathbf D$ is a submatrix of $\mathbf D'$.
Now:
:$\paren {a' b'}_{i j} = \begin{cases} \ds \sum_{r \mathop = 1}^{n + 1} \mathbf a'_{i r} b'_{r j} & : i < n + 1 \land j < n + 1 \\ 0 & : i = n + 1 \lor j = n + 1 \end{cases}$
But:
and so:
Thus it is seen that:
:$\exists \mathbf A', \mathbf B' \in \MM_{n + 1 \times n + 1}: \mathbf A' \mathbf B' \ne \mathbf B' \mathbf A'$
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\exists \mathbf A, \mathbf B \in \map {\MM_R} n: \mathbf {A B} \ne \mathbf {B A}$
and by definition (conventional) matrix multiplication over $\map {\MM_R} n$ is not commutative.
== Examples ==
== Also see ==
* Definition:Pre-Multiplication
* Definition:Post-Multiplication
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.2$ Addition and multiplication of matrices
Category:Conventional Matrix Multiplication
Category:Examples of Commutative Operations
Category:Proofs by Induction
Category:Matrix Multiplication is not Commutative