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:
 * $\mathbf {0 A}$ is defined

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

or both.

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

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

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 $\mathbf {0 A}$ be defined.

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

Thus we have:

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

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

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

Thus we have:

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

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