In Exercises 25-28, determine if the specified linear transformation is (a) one-to-one and (b) onto. Justify each answer.

26. The transformation in Exercise 2.

Write the transformation in Exercise 2.

\(T:{\mathbb{R}^3} \to {\mathbb{R}^2}\), \(T\left( {{e_1}} \right) = \left( {1,3} \right)\), \(T\left( {{e_2}} \right) = \left( {4, - 7} \right)\), and \(T\left( {{e_3}} \right) = \left( { - 5,4} \right)\), where \({e_1},{e_2},{e_3}\) are the columns of a \(3 \times 3\) identity matrix.

Write the vector into the columns of a matrix \(A\).

\(\begin{array}{c}T\left( x \right) = \left( {\begin{array}{*{20}{c}}{T\left( {{e_1}} \right)}&{T\left( {{e_1}} \right)}&{T\left( {{e_1}} \right)}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}1&4&{ - 5}\\3&{ - 7}&4\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right)\\ = Ax\end{array}\)

At row 2, multiply row 1 by 3 and subtract it from row 2.

\(\left( {\begin{array}{*{20}{c}}1&4&{ - 5}\\0&{ - 19}&{19}\end{array}} \right)\)

Theorem 12states that let \(T:{\mathbb{R}^n} \to {\mathbb{R}^m}\) be a linear transformation, and let \(A\) be the standard matrix \(T\) then:

1. \(T\)maps \({\mathbb{R}^n}\) onto \({\mathbb{R}^m}\) if and only if the columns of \(A\) span \({\mathbb{R}^m}\).
2. \(T\)is one-to-one if and only if the columns of \(A\) are linearly independent.

Matrix \(A\) has more columns than rows; so the columns of \(A\) are linearly dependent.\(T\)is not one-to-one according to theorem 12 since the rows of \(A\) span \({\mathbb{R}^2}\). Therefore, \(T\) maps \({\mathbb{R}^3}\), according to theorem 12.

Thus, the specified linear transformation is onto.