Zero Matrix is Zero for Matrix Multiplication

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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:

\(\ds \mathbf 0 \mathbf A\) \(=\) \(\ds \mathbf C\)
\(\ds \sqbrk 0_{r m} \sqbrk a_{m n}\) \(=\) \(\ds \sqbrk c_{r n}\) 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
\(\ds \) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \mathbf 0 \mathbf A\) \(=\) \(\ds \sqbrk 0_{r n}\)

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:

\(\ds \mathbf A \mathbf 0\) \(=\) \(\ds \mathbf C\)
\(\ds \sqbrk a_{m n} \sqbrk 0_{n s}\) \(=\) \(\ds \sqbrk c_{m s}\) 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
\(\ds \) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \mathbf 0 \mathbf A\) \(=\) \(\ds \sqbrk 0_{m s}\)

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

$\Box$


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

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

$\blacksquare$


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