Order 2 Square Matrices with Zero Diagonals do not form Ring

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Theorem

Let $S$ be the set of square matrices of order $2$ whose diagonal elements are zero.


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


Note that $\times$ denotes conventional matrix multiplication.


Proof

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 $x = \begin {pmatrix} 0 & 1 \\ 1 & 0 \end {pmatrix}$ and $y = \begin {pmatrix} 0 & 2 \\ 2 & 0 \end {pmatrix}$.

Both $x$ and $y$ are in $S$, as both are square matrices of order $2$ whose diagonal elements are zero.

But then:

$x y = \begin {pmatrix} 0 & 1 \\ 1 & 0 \end {pmatrix} \begin {pmatrix} 0 & 2 \\ 2 & 0 \end {pmatrix} = \begin {pmatrix} 2 & 0 \\ 0 & 2 \end {pmatrix}$

which is a square matrix of order $2$ whose diagonal elements are not zero.

Hence $x y \notin S$ and so $\struct {S, \times}$ is not closed.

The result follows.

$\blacksquare$


Sources

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