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:
\(\ds \mathbf {A 0}\) | \(=\) | \(\ds \sqbrk 0_{m m}\) | ||||||||||||
\(\ds \mathbf {0 A}\) | \(=\) | \(\ds \sqbrk 0_{n n}\) |
$\blacksquare$
Sources
- 1998: Richard Kaye and Robert Wilson: Linear Algebra ... (previous) ... (next): Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.2$ Addition and multiplication of matrices: $9$