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 \(T\left( {{x_1},{x_2}} \right)\).

36.Let \(T:{\mathbb{R}^3} \to {\mathbb{R}^3}\) be the transformation that projects each vector \({\bf{x}} = \left( {{x_1},{x_2},{x_3}} \right)\) onto the plane \({x_2} = 0\), so \(T\left( {\bf{x}} \right) = T\left( {{x_1},0,{x_3}} \right)\). Show that T is a linear transformation.

The transformation \(T\) is said to be linear if all vectors \({\bf{u}}\) in \({\mathbb{R}^n}\) and all scalars \(c\) and \(d\) are represented in the domain \(T\)as shown below:

• \(T\left( {c{\bf{u}}} \right) = cT\left( {\bf{u}} \right)\)
• \(T\left( {c{\bf{u}} + d{\bf{v}}} \right) = cT\left( {\bf{u}} \right) + dT\left( {\bf{v}} \right)\)

Let\({\bf{u}} = \left( {{u_1},{u_2},{u_3}} \right)\), and\({\bf{v}} = \left( {{v_1},{v_2},{v_3}} \right)\).

Substitute\({\bf{u}} = \left( {{u_1},{u_2},{u_3}} \right)\), and\({\bf{v}} = \left( {{v_1},{v_2},{v_3}} \right)\)in\(c{\bf{u}} + d{\bf{v}}\) as shown below:

\(\begin{aligned}{c}c{\bf{u}} + d{\bf{v}} &= c\left( {{u_1},{u_2},{u_3}} \right) + d\left( {{v_1},{v_2},{v_3}} \right)\\ &= \left( {c{u_1},c{u_2},c{u_3}} \right) + \left( {d{v_1},d{v_2},d{v_3}} \right)\\ &= \left( {c{u_1} + d{v_1},c{u_2} + d{v_2},c{u_3} + d{v_3}} \right)\end{aligned}\)

As the vector is \(c{\bf{u}} + d{\bf{v}} = \left( {c{u_1} + d{v_1},c{u_2} + d{v_2},c{u_3} + d{v_3}} \right)\); apply the transformation by using the concept that for\({\bf{x}} = \left( {{x_1},{x_2},{x_3}} \right)\), the transformation is\(T\left( {\bf{x}} \right) = T\left( {{x_1},0,{x_3}} \right)\).

\(\begin{aligned}{c}T\left( {c{\bf{u}} + d{\bf{v}}} \right) &= T\left( {c{u_1} + d{v_1},c{u_2} + d{v_2},c{u_3} + d{v_3}} \right)\\ &= \left( {c{u_1} + d{v_1},0,c{u_3} + d{v_3}} \right)\\ &= \left( {c{u_1} + d{v_1},0,c{u_3} + d{v_3}} \right)\\ &= \left( {c{u_1},0,c{u_3}} \right) + \left( {d{v_1},0,d{v_3}} \right)\end{aligned}\)

Simplify further.

\(\begin{aligned}{c}T\left( {c{\bf{u}} + d{\bf{v}}} \right) &= c\left( {{u_1},0,{u_3}} \right) + d\left( {{v_1},0,{v_3}} \right)\\ &= cT\left( {\bf{u}} \right) + dT\left( {\bf{v}} \right)\end{aligned}\)

Since \(T\left( {c{\bf{u}} + d{\bf{v}}} \right) = cT\left( {\bf{u}} \right) + dT\left( {\bf{v}} \right)\), \(T\) is a linear transformation.