# 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:

 $\ds \mathbf {0 A}$ $=$ $\ds \mathbf C$ $\ds \sqbrk 0_{r m} \sqbrk a_{m n}$ $=$ $\ds \sqbrk c_{r n}$ Definition of $\mathbf 0$ and $\mathbf A$ $\ds \leadsto \ \$ $\ds \forall i \in \closedint 1 r, j \in \closedint 1 n: \,$ $\ds c_{i j}$ $=$ $\ds \sum_{k \mathop = 1}^m 0_{i k} \times a_{k j}$ Definition of Matrix Product (Conventional) $\ds$ $=$ $\ds \sum_{k \mathop = 1}^m 0$ Definition of Zero Matrix $\ds$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds \mathbf {0 A}$ $=$ $\ds \sqbrk 0_{r n}$

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:

 $\ds \mathbf {A 0}$ $=$ $\ds \mathbf C$ $\ds \sqbrk a_{m n} \sqbrk 0_{n s}$ $=$ $\ds \sqbrk c_{m s}$ Definition of $\mathbf A$ and $\mathbf 0$ $\ds \leadsto \ \$ $\ds \forall i \in \closedint 1 m, j \in \closedint 1 s: \,$ $\ds c_{i j}$ $=$ $\ds \sum_{k \mathop = 1}^n a_{i k} \times 0_{k j}$ Definition of Matrix Product (Conventional) $\ds$ $=$ $\ds \sum_{k \mathop = 1}^n 0$ Definition of Zero Matrix $\ds$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds \mathbf {0 A}$ $=$ $\ds \sqbrk 0_{m s}$

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

$\Box$

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

 $\ds \mathbf {A 0}$ $=$ $\ds \sqbrk 0_{m m}$ $\ds \mathbf {0 A}$ $=$ $\ds \sqbrk 0_{n n}$

$\blacksquare$