Zero Matrices
Zero Matrices
Definition An $m \times n$ matrix $A$ is a Zero Matrix if all entries in the matrix are $0$, that is $a_{ij} = 0$ for all $1 ≤ i ≤ m$ and $1 ≤ j ≤ n$, $i, j \in \mathbb{N}$. |
The definition of a zero matrix is pretty self explanatory. For example, if $A$ is a $2 \times 3$ zero matrix would look like this $A = \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}$.
Often times zero matrices of size $m \times n$ are denoted simply by $0_{m \times n }$.
We will now look at a rather simple theorem regarding various operations with the zero matrix.
Theorem 1: Let $A$ be an $m \times n$ matrix and let $0$ be the $m \times n$ zero matrix. Then: a) $A + 0 = 0 + A = A$. b) $0 - A = -A$. c) $A - A = 0$. d) $A \cdot 0 = 0$. |
- Proof of (a): Suppose that $A$ and $0$ are matrices of size $m \times n$. Thus:
\begin{align} \quad A + 0 = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{m1} & a_{m2} & \cdots& a_{mn} \end{bmatrix} + \begin{bmatrix} 0 & 0 & \cdots & 0\\ 0 & 0 & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & 0 \end{bmatrix}_{m \times n} = \begin{bmatrix} a_{11}+0 & a_{12}+0 & \cdots & a_{1n}+0\\ a_{21}+0 & a_{22}+0 & \cdots & a_{2n}+0\\ \vdots & \vdots & \ddots & \vdots\\ a_{m1}+0 & a_{m2}+0 & \cdots& a_{mn}+0 \end{bmatrix} = A \\ \blacksquare \end{align}
- Proof of (b): We know that by (a) that $A + 0 = 0 + A = A$. It follows that $0 - A = 0 + (-A) = -A$. $\blacksquare$
- Proof of (c): Suppose $A$ is a matrix of size $m \times n$, then:
\begin{align} \quad A - A = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{m1} & a_{m2} & \cdots& a_{mn} \end{bmatrix} - \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{m1} & a_{m2} & \cdots& a_{mn} \end{bmatrix} = \begin{bmatrix} a_{11} - a_{11} & a_{12} - a_{12} & \cdots & a_{1n} - a_{1n}\\ a_{21} - a_{21} & a_{22} - a_{22} & \cdots & a_{2n} - a_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{m1} - a_{m1} & a_{m2} - a_{m2} & \cdots& a_{mn} - a_{mn} \end{bmatrix} = \begin{bmatrix} 0 & 0 & \cdots & 0\\ 0 & 0 & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & 0 \end{bmatrix}_{m \times n} = 0 \\ \blacksquare \end{align}
- Proof of (d): Recall that any entry of a product of two matrices can be determined by the formula $(AB)_{ij} = a_{i1}b_{1j} + a_{i2}b_{2j} + ... + a_{ir}b_{rj}$. Let $A$ be an $m \times r$ matrix and let $0$ be an $r \times n$ matrix. By definition, all entries of $B$ are zeroes, hence $b_{ij} = 0$ for all $i$ and $j$, and thus $(A \cdot 0)_{ij} = a_{i1}\cdot 0 + a_{i2}\cdot 0 + ... + a_{ir} \cdot 0 = 0_{m \times n}$. $\blacksquare$