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Section 2.3 Scalar multiplication

Definition 2.3.1. Scalars.

A scalar is a real number.

We next define the multiplication of a scalar and a matrix.

Definition 2.3.2. Scalar multiplication.

If \(A=[a_{i,j}]\) is a matrix and \(r\) is a scalar, then the matrix \(C=[c_{i,j}]=rA\) is defined by

\begin{equation*} c_{i,j}=ra_{i,j} \end{equation*}

In other words, every entry of the matrix \(C\) is multiplied by \(r\text{.}\)

If

\begin{equation*} A= \begin{bmatrix} 1 \amp 2 \amp 3\\ 4 \amp 5 \amp 6 \end{bmatrix} \text{ and } r=2 \end{equation*}

then

\begin{equation*} rA=2A= \begin{bmatrix} 2 \amp 4 \amp 6\\ 8 \amp 10 \amp 12 \end{bmatrix}. \end{equation*}

In short, the product \(rA\) is computed by multiplying every entry of \(A\) by \(r\text{.}\)

  • By the definition of scalar multiplication, if \(A\) is an \(m\times n\) matrix, then \(rA\) is an \(m\times n\) matrix also.

  • We use \(A=[a_{i,j}]\) and \(B=[b_{i,j}]\text{.}\) Then the \(i\)-\(j\) entry of \(r(A+B)\) is \(r(a_{i,j}+b_{i,j})\) while he \(i\)-\(j\) entry of \(rA+rB\) is \(ra_{i,j}+rb_{i,j}\text{.}\) Hence the for the equality to be valid, we need \(r(a_{i,j}+b_{i,j})=ra_{i,j}+rb_{i,j}\text{,}\) which is the distributive law for real numbers. (see properties of real numbers in Subsection 7.1.2.)

  • We use \(A=[a_{i,j}].\) The \(i\)-\(j\) entry of \((r+s)A\) is \((r+s)a_{i,j}\) while the \(i\)-\(j\) entry of \(rA+sA\) is \(ra_{i,j}+sa_{i,j}\text{.}\) Hence the matrix equation is valid if \((r+s)a_{i,j}=ra_{i,j}+sa_{i,j}\text{.}\) Since this equation is the distributive law for real numbers, the validity is clear.

  • We use \(A=[a_{i,j}].\) The \(i\)-\(j\) entry of \((rs)A\) is \((rs)a_{i,j}\) while the \(i\)-\(j\) entry of \(r(sA)\) is \(r(sa_{i,j}).\) Hence the matrix equation is valid if \((rs)a_{i,j}=r(sa_{i,j}).\) Since this is the associative law for real numbers, the result is clear.

  • We use \(A=[a_{i,j}].\) The \(i\)-\(j\) entry of \(1A\) is \(1a_{i,j}=a_{i,j}\text{,}\) and so \(1A=A.\)

Linear combinations.

Matrix addition and scalar multiplication are both used to compute linear combinations.

Suppose \(A=\begin{bmatrix} 1\amp 2\\3\amp 4 \end{bmatrix}\) and \(B=\begin{bmatrix} -1\amp 0\\2\amp 1 \end{bmatrix}\text{.}\) Then the expression \(2A+3B\) makes sense and can be evaluated as

\begin{equation*} \begin{array}{rl} 2A+3B \amp =2 \begin{bmatrix} 1\amp 2\\3\amp 4 \end{bmatrix} +3\begin{bmatrix} -1\amp 0\\2\amp 1 \end{bmatrix}\\ \amp =\begin{bmatrix} 2\amp 4\\6\amp 8 \end{bmatrix} +\begin{bmatrix} -3\amp 0\\6\amp 3 \end{bmatrix}\\ \amp =\begin{bmatrix} -1\amp 4\\ 12\amp 11 \end{bmatrix} \end{array}\text{.} \end{equation*}
Definition 2.3.6. Linear combination of two matrices.

If \(A\) and \(B\) are matrices conformable for addition, and \(r\) and \(s\) are scalars, then the matrix of the form \[ rA+sB \] is called a linear combination of \(A\) and \(B\text{.}\)

The concept can be applied easily to more than two matrices.

Definition 2.3.7. Linear combination of matrices.

If \(A_1,A_2,\ldots,A_n\) are matrices conformable for addition, then, for any choice of scalars \(r_1,r_2,\ldots,r_n\text{,}\) the matrix \[ r_1A_1+r_2A_2+\cdots+r_nA_n \] is called a linear combination of \(A_1,A_2,\ldots,A_n\).

Exercises Scalar multiplication exercises

1.

Let \(A= \begin{bmatrix} 1\amp2 \amp3\\ 3\amp2\amp1 \end{bmatrix}\) Evaluate

  • \(\displaystyle 2A\)

  • \(\displaystyle -2A\)

  • \(\displaystyle \frac12 A\)

  • \(\displaystyle (-1) A\)

  • \(\displaystyle 0A\)

Solution.

  • \(\displaystyle 2A= \begin{bmatrix} 2\amp 4\amp 6\\ 6\amp 4\amp 2 \end{bmatrix}\)

  • \(\displaystyle -2A= \begin{bmatrix} -2\amp -4\amp -6\\ -6\amp -4\amp -2 \end{bmatrix}\)

  • \(\displaystyle \frac12 A= \begin{bmatrix} \frac12 \amp 1\amp \frac32\\ \frac32 \amp 1\amp \frac12 \end{bmatrix}\)

  • \(\displaystyle (-1) A= \begin{bmatrix} -1\amp-2 \amp-3\\ -3\amp-2\amp-1 \end{bmatrix}\)

  • \(\displaystyle 0A= \begin{bmatrix} 0\amp0 \amp0\\ 0\amp0\amp0 \end{bmatrix}\)

2.

Recall the that zero matrix \(\vec0\) has every entry equal to zero. Show that \(A+(-1)A=\vec 0\text{.}\)

Solution.

The \(i\)-\(j\) entry of \(A+(-1)A\) is \(a_{i,j}+(-1)a_{i,j}=0\) and so \(A+(-1)A=\vec 0\text{.}\)

3.

Let \(A\) be a any matrix and \(r\) any scalar. Show that \(0A=r\vec 0\) where \(\vec 0\) is the zero matrix with the same size as \(A\text{.}\)

Solution.

The \(i\)-\(j\) entry of \(0A\) is \(0a_{i,j}=0\) and the \(i\)-\(j\) entry of \(r\vec0\) is \(r0=0\text{.}\) Hence \(0A=\vec0=r\vec0\text{.}\)