Section 3.3 Scalar multiplication
Definition 3.3.1. Scalars.
A scalar is a real number.
We next define the multiplication of a scalar and a matrix.
Definition 3.3.2. 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
In other words, every entry of the matrix \(C\) is multiplied by \(r\text{.}\)
Example 3.3.3. Scalar multiplication of a matrix.
If
then
In short, the product \(rA\) is computed by multiplying every entry of \(A\) by \(r\text{.}\)
Theorem 3.3.4. Properties of scalar multiplication.
Suppose that \(A\) and \(B\) are matrices of the same size, and \(r\) and \(s\) are scalars, then
If \(A\) is an \(m\times n\) matrix, then \(rA\) is also \(m\times n\text{.}\)
\(\displaystyle r(A+B)=rA+rB\)
\(\displaystyle (r+s)A=rA+sA\)
\(\displaystyle (rs)A=r(sA)\)
\(\displaystyle 1A=A\)
Proof.
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 8.1.2.)
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.
Example 3.3.5. A linear combination.
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
Definition 3.3.6. 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 3.3.7. 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
\(\displaystyle 2A\)
\(\displaystyle -2A\)
\(\displaystyle \frac12 A\)
\(\displaystyle (-1) A\)
\(\displaystyle 0A\)
\(\displaystyle 2A= \begin{bmatrix} 2\amp 4\amp 6\\ 6\amp 4\amp 2 \end{bmatrix}\)
\(\displaystyle -2A= \begin{bmatrix} -2\amp -4\amp -6\\ -6\amp -4\amp -2 \end{bmatrix}\)
\(\displaystyle \frac12 A= \begin{bmatrix} \frac12 \amp 1\amp \frac32\\ \frac32 \amp 1\amp \frac12 \end{bmatrix}\)
\(\displaystyle (-1) A= \begin{bmatrix} -1\amp-2 \amp-3\\ -3\amp-2\amp-1 \end{bmatrix}\)
\(\displaystyle 0A= \begin{bmatrix} 0\amp0 \amp0\\ 0\amp0\amp0 \end{bmatrix}\)
2.
Recall the that zero matrix \(\vec0\) has every entry equal to zero. Show that \(A+(-1)A=\vec 0\text{.}\)
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 \(A\text{.}\)
The \(i\)-\(j\) entry of \(0A\) is \(0a_{i,j}=0\) and the \(i\)-\(j\) entry of \(r\vec0\) is \(r0=0\text{.}\) Hence \(0A=\vec0=r\vec0\text{.}\)