Let \(B = \left\{ {{{\bf{b}}_1},{{\bf{b}}_2},{{\bf{b}}_3}} \right\}\)be a basis for a vector space \(V\) and\(T:V \to {\mathbb{R}^2}\) be a linear transformation with the property that

\(T\left( {{x_1}{{\bf{b}}_1} + {x_2}{{\bf{b}}_2} + {x_3}{{\bf{b}}_3}} \right) = \left( {\begin{aligned}{2{x_1} - 4{x_2} + 5{x_3}}\\{ - {x_2} + 3{x_3}}\end{aligned}} \right)\)

Find the matrix for \(T\) relative to \(B\) and the standard basis for \({\mathbb{R}^2}\).

The standard basis for \({\mathbb{R}^2}\) is given by \({{\bf{e}}_1},{{\bf{e}}_2}\).

\({{\bf{e}}_1} = \left( {\begin{aligned}1\\0\end{aligned}} \right)\), \({{\bf{e}}_2} = \left( {\begin{aligned}0\\1\end{aligned}} \right)\)

Let \(\varepsilon = \left\{ {{{\bf{e}}_1},{{\bf{e}}_2}} \right\}\).

Find \({\left( {T\left( {{{\bf{b}}_1}} \right)} \right)_\varepsilon }\) by using \(T\left( {{x_1}{{\bf{b}}_1},{x_2}{{\bf{b}}_2},{x_3}{{\bf{b}}_3}} \right) = \left( {\begin{aligned}{2{x_1} - 4{x_2} + 5{x_3}}\\{ - {x_2} + 3{x_3}}\end{aligned}} \right)\).

\(\begin{aligned}{c}{\left( {T\left( {{{\bf{b}}_1}} \right)} \right)_\varepsilon } &= T\left( {1{{\bf{b}}_1} + 0{{\bf{b}}_2} + 0{{\bf{b}}_3}} \right)\\ &= \left( {\begin{aligned}2\\0\end{aligned}} \right)\end{aligned}\)

Find \({\left( {T\left( {{{\bf{b}}_2}} \right)} \right)_\varepsilon }\) by using \(T\left( {{x_1}{{\bf{b}}_1},{x_2}{{\bf{b}}_2},{x_3}{{\bf{b}}_3}} \right) = \left( {\begin{aligned}{2{x_1} - 4{x_2} + 5{x_3}}\\{ - {x_2} + 3{x_3}}\end{aligned}} \right)\).

\(\begin{aligned}{c}{\left( {T\left( {{{\bf{b}}_2}} \right)} \right)_\varepsilon } &= T\left( {0{{\bf{b}}_1} + 1{{\bf{b}}_2} + 0{{\bf{b}}_3}} \right)\\ &= \left( {\begin{aligned}{ - 4}\\{ - 1}\end{aligned}} \right)\end{aligned}\)

Find \({\left( {T\left( {{{\bf{b}}_3}} \right)} \right)_\varepsilon }\) by using \(T\left( {{x_1}{{\bf{b}}_1},{x_2}{{\bf{b}}_2},{x_3}{{\bf{b}}_3}} \right) = \left( {\begin{aligned}{2{x_1} - 4{x_2} + 5{x_3}}\\{ - {x_2} + 3{x_3}}\end{aligned}} \right)\).

\(\begin{aligned}{c}{\left( {T\left( {{{\bf{b}}_3}} \right)} \right)_\varepsilon } &= T\left( {0{{\bf{b}}_1} + 0{{\bf{b}}_2} + 1{{\bf{b}}_3}} \right)\\ &= \left( {\begin{aligned}5\\3\end{aligned}} \right)\end{aligned}\)

A matrix associated with a linear transformation \(T\) for \(V\) and \(W\) is given by \({\left( {T\left( {\bf{x}} \right)} \right)_C} = \left( {\begin{aligned}{{{\left( {T\left( {{{\bf{b}}_1}} \right)} \right)}_C}}&{{{\left( {T\left( {{{\bf{b}}_2}} \right)} \right)}_C}}& \cdots &{{{\left( {T\left( {{{\bf{b}}_n}} \right)} \right)}_C}}\end{aligned}} \right)\), where \(V\) and \(W\) are \(n\) and \(m\)-dimensional subspaces respectively, and \(B\), and\(C\) are the bases for \(V\), and \(W\).

Form a matrix \(T\) for the obtained vectors in step 2, by using the formula \({\left( {T\left( {\bf{x}} \right)} \right)_C} = \left( {\begin{aligned}{{{\left( {T\left( {{{\bf{b}}_1}} \right)} \right)}_C}}&{{{\left( {T\left( {{{\bf{b}}_2}} \right)} \right)}_C}}& \cdots &{{{\left( {T\left( {{{\bf{b}}_n}} \right)} \right)}_C}}\end{aligned}} \right)\), where \(n = 3\).

\(\begin{aligned}{\left( {T\left( {\bf{x}} \right)} \right)_\varepsilon } = \left( {\begin{aligned}{{{\left( {T\left( {{{\bf{b}}_1}} \right)} \right)}_\varepsilon }}&{{{\left( {T\left( {{{\bf{b}}_2}} \right)} \right)}_\varepsilon }}&{{{\left( {T\left( {{{\bf{b}}_3}} \right)} \right)}_\varepsilon }}\end{aligned}} \right)\\ = \left( {\begin{aligned}2&{}&{ - 4}&{}&5\\0&{}&{ - 1}&{}&3\end{aligned}} \right)\end{aligned}\)

So, the required matrix is \(\left( {\begin{aligned}2&{}&{ - 4}&{}&5\\0&{}&{ - 1}&{}&3\end{aligned}} \right)\).