Scalar Multiples of Matrices

Definition: A scalar is a quantity that has a magnitude (size or length) but no direction.

Unless otherwise specified, scalars in the context of Linear Algebra will be a real number $k \in \mathbb{R}$. We are now ready to define scalar multiplication on a matrix.

Definition: If $A$ is an $m \times n$ matrix and $k \in \mathbb{R}$ a scalar, then the scalar multiple of $A$ by $k$ denoted $kA$ is an $m \times n$ matrix, all of whose entries are multiplied by $k$.

Determining a scalar multiple of a matrix is easy. For example, consider the matrix $A = A = \begin{bmatrix}1 & 2 & 3 & 4\\ 5 & 6 & 7 & 8 \end{bmatrix}$. If we wanted to figure out what the matrix $2A$ is, we would just take every entry in $A$ and multiply it by $2$ to get $2A = \begin{bmatrix}2\cdot1 & 2\cdot2 & 2\cdot3 & 2\cdot4\\ 2\cdot5 & 2\cdot6 & 2\cdot7 & 2\cdot8\end{bmatrix} = \begin{bmatrix}2 & 4 & 6& 8\\ 10 & 12 & 14 & 16\end{bmatrix}$

In general, if we have an $m \times n$ matrix $A = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & & a_{2n}\\ \vdots & & \ddots & \vdots\\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix}$ and a scalar $k \in \mathbb{R}$ then:

(1)

\begin{align} kA = k \: \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} ka_{11} & ka_{12} & \cdots & ka_{1n}\\ ka_{21} & ka_{22} & \cdots & ka_{2n} \\ \vdots & \vdots & \ddots & \vdots\\ ka_{m1} & ka_{m2} & \cdots & ka_{mn} \end{bmatrix} \end{align}

We will now look at some properties of scalar multiples of matrices in the following theorem.

Theorem 1: Let $A$ and $B$ be $m \times n$ matrices, and let $k, l \in \mathbb{R}$ be scalars. Then: a) $k(A + B) = kA + kB$. b) $k(A - B) = kA - kB$. c) $(k + l)A = kA + lA$. d) $(k - l)A = kA - lA$. e) $k(lA) = (kl) A$.

We will prove (a) and (c) and leave the rest of the proofs to the reader as they follow the same format.

• Proof of (a):

(2)

\begin{align} k(A + B) = k\begin{bmatrix} a_{11} + b_{11} & a_{12} + b_{12} & \cdots & a_{1n} + b_{1n}\\ a_{21} + b_{21} & a_{22} + b_{22} & \cdots & a_{2n} + b_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{m1} + b_{m1} & a_{m2} + b_{m2} & \cdots& a_{mn} + b_{mn} \end{bmatrix} = \begin{bmatrix} ka_{11} + kb_{11} & ka_{12} + kb_{12} & \cdots & ka_{1n} + kb_{1n}\\ ka_{21} + kb_{21} & ka_{22} + kb_{22} & \cdots & ka_{2n} + kb_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ ka_{m1} + kb_{m1} & ka_{m2} + kb_{m2} & \cdots& ka_{mn} + kb_{mn} \end{bmatrix}\\ = k\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} + k\begin{bmatrix} b_{11} & b_{12} & \cdots & b_{1n}\\ b_{21} & b_{22} & & b_{2n}\\ \vdots & & \ddots & \vdots\\ b_{m1} & b_{m2} & \cdots & b_{mn} \end{bmatrix} = kA + kB \\ \blacksquare \end{align}

• Proof of (c):

(3)

\begin{align} \quad (k + l)A = \begin{bmatrix} (k+l)a_{11} & (k+l)a_{12} & \cdots & (k+l)a_{1n}\\ (k+l)a_{21} & (k+l)a_{22} & \cdots & (k+l)a_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ (k+l)a_{m1} & (k+l)a_{m2} & \cdots& (k+l)a_{mn}\end{bmatrix} = \begin{bmatrix} ka_{11} + la_{11} & ka_{12} + lka_{12}& \cdots & ka_{1n} + l\\ ka_{21} + la_{21} & ka_{22} + la_{22} & \cdots & ka_{2n} + la_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ ka_{m1} + la_{m1} & ka_{m2} + la_{m2} & \cdots& ka_{mn} + la_{mn} \end{bmatrix} = kA + lA \\ \blacksquare \end{align}

Example 1

Given the matrix $A = \begin{bmatrix}2 & 4\\ 1 & 0\\ 0 & 3\\ -3 & 2\\ 1 & 3\end{bmatrix}$, find the matrix $-3A$.

To determine $-3A$, we will multiply every entry in A by -3 to obtain $-3A = \begin{bmatrix} -6 & -12\\ -3 & 0\\ 0 & -9\\ 9 & -6\\ -3 & -9 \end{bmatrix}$.