Zero Matrix is Zero for Matrix Multiplication

Theorem
Let $\struct {R, +, \times}$ be a ring.

Let $\mathbf A$ be a matrix over $R$ of order $m \times n$

Let $\mathbf 0$ be a zero matrix whose order is such that either:
 * $\mathbf 0 \mathbf A$ is defined

or:
 * $\mathbf A \mathbf 0$ is defined

or both.

Then:
 * $\mathbf 0 \mathbf A = \mathbf 0$

or:
 * $\mathbf A \mathbf 0 = \mathbf 0$

whenever they are defined.

The order of $\mathbf 0$ will be according to the orders of the factor matrices.

Proof
Let $\mathbf A = \sqbrk a_{m n}$ be matrices.

Let $\mathbf 0 \mathbf A$ be defined.

Then $\mathbf 0$ is of order $r \times m$ for $r \in \Z_{>0}$.

Thus we have:

Hence $\mathbf 0 \mathbf A$ is the Zero Matrix of order $r \times n$.

Let $\mathbf A \mathbf 0$ be defined.

Then $\mathbf 0$ is of order $n \times s$ for $s \in \Z_{>0}$.

Thus we have:

Hence $\mathbf A \mathbf 0$ is the Zero Matrix of order $m \times s$.

If $\mathbf 0$ is of order $n \times m$,then both $\mathbf A \mathbf 0$ and $\mathbf 0 \mathbf A$ are defined, and: