In Exercises 32–36, column vectors are written as rows, such as \({\bf{x}} = \left( {{x_1},{x_2}} \right)\), and \(T\left( {\bf{x}} \right)\) is written as \(T\left( {{x_1},{x_2}} \right)\).

32. Show that the transformation \(T\) defined by \(T\left( {{x_1},{x_2}} \right) = \left( {4{x_1} - 2{x_2},3\left| {{x_2}} \right|} \right)\)is not linear.

The transformation \(T\) is said to be linear if all vectors \({\bf{u}}\) in \({\mathbb{R}^n}\) and all scalars \(c\) are represented in the domain \(T\), such that \(T\left( {c{\bf{u}}} \right) = cT\left( {\bf{u}} \right)\).

Assume that the vector is\({\bf{u}} = \left( {0,1} \right)\), and scalar is\(c = - 1\).

Take transformation on both sides for\({\bf{u}} = \left( {0,1} \right)\).

\(T\left( {\bf{u}} \right) = T\left( {0,1} \right)\)

Substitute 0 for\({x_1}\), and 1 for\({x_2}\)in the transformation\(T\left( {{x_1},{x_2}} \right) = \left( {4{x_1} - 2{x_2},3\left| {{x_2}} \right|} \right)\).

\(\begin{aligned}{c}T\left( {0,1} \right) &= \left( {4\left( 0 \right) - 2\left( 1 \right),3\left| 1 \right|} \right)\\ &= \left( { - 2,3} \right)\end{aligned}\)

Thus, \(T\left( {\bf{u}} \right) = \left( { - 2,3} \right)\).

Use the scalar\(c = - 1\)and the scalar multiplication to obtain the transformation\(T\left( {c{\bf{u}}} \right)\).

\(\begin{aligned}{c}T\left( {c{\bf{u}}} \right) &= T\left( {\left( { - 1} \right) \cdot \left( {0,1} \right)} \right)\\ &= T\left( {\left( { - 1} \right)\left( 0 \right),\left( { - 1} \right)\left( 1 \right)} \right)\\ &= T\left( {0, - 1} \right)\end{aligned}\)

Solve \(T\left( {0, - 1} \right)\) by using \(T\left( {{x_1},{x_2}} \right) = \left( {4{x_1} - 2{x_2},3\left| {{x_2}} \right|} \right)\).

\(\begin{aligned}{c}T\left( {0, - 1} \right) &= \left( {4\left( 0 \right) - 2\left( { - 1} \right),3\left| { - 1} \right|} \right)\\ &= \left( {2,3} \right)\end{aligned}\)

Obtain the transformation \(cT\left( {\bf{u}} \right)\).

\(\begin{aligned}{c}cT\left( {\bf{u}} \right) &= \left( { - 1} \right)T\left( {\left( {0,1} \right)} \right)\\ &= \left( { - 1} \right)\left( {4\left( 0 \right) - 2\left( 1 \right),3\left| 1 \right|} \right)\\ &= \left( { - 1} \right)\left( { - 2,3} \right)\end{aligned}\)

Solve further.

\(\begin{aligned}{c}cT\left( {\bf{u}} \right) &= \left( {\left( { - 1} \right)\left( { - 2} \right),\left( { - 1} \right)\left( 3 \right)} \right)\\ &= \left( {2, - 3} \right)\end{aligned}\)

It is observed that\(T\left( {c{\bf{u}}} \right) = \left( {2,3} \right)\)and \(cT\left( {\bf{u}} \right) = \left( {2, - 3} \right)\). This means that \(T\left( {c{\bf{u}}} \right) \ne cT\left( {\bf{u}} \right)\).

Thus, the transformation \(T\left( {\bf{x}} \right)\) is not linear.