Set of Order m times n Matrices does not form Ring

Theorem

Let $m, n \in \N_{>0}$ be non-zero natural numbers such that $m > n$.

Then the algebraic structure $\struct {S, +, \times}$ is not a ring.

For $\struct {S, +, \times}$ to be a ring, it is a necessary condition that $\struct {S, \times}$ is a semigroup.

For $\struct {S, \times}$ to be a semigroup, it is a necessary condition that $\struct {S, \times}$ is closed.

That is:

$\forall x, y \in S: x \times y \in S$

Let $\mathbf A$ and $\mathbf B$ be elements of $S$.

The matrix multiplication operation is defined on $\mathbf A$ and $\mathbf B$ if and only if $\mathbf A$ is of order $m \times p$ and $\mathbf A$ is of order $p \times n$, for some $m, n, p \in \N_{>0}$.

That is, the second dimension of $\mathbf A$ must be equal to the first dimension of $\mathbf B$.

But in this case, the second dimension of $\mathbf A$ is $n$, and the first dimension of $\mathbf B$ is $m$.

As we have been given that $m \ne n$, it follows that matrix multiplication operation is not defined on $S$.

Hence the result.

$\blacksquare$

1970: B. Hartley and T.O. Hawkes: Rings, Modules and Linear Algebra ... (previous) ... (next): Chapter $1$: Rings - Definitions and Examples: $2$: Some examples of rings: Some 'non-examples': $\text {(e)}$