91Ó°ÊÓ

Write the transformation in Exercise 19.

\(T\left( {{x_1},{x_2},{x_3}} \right) = \left( {{x_1} - 5{x_2} + 4{x_3},{x_2} - 6{x_3}} \right)\)

Write the transformation \(T\left( x \right)\) and \(x\) into the column vectors of \(A\).

\(\begin{array}{c}T\left( x \right) = \left( {\begin{array}{*{20}{c}}{{x_1} - 5{x_2} + 4{x_3}}\\{{x_2} - 6{x_3}}\end{array}} \right))\\ = \left( A \right))\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right))\\ = \left( {\begin{array}{*{20}{c}}1&{ - 5}&4\\0&1&{ - 6}\end{array}} \right))\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right))\end{array}\))

Theorem 12 states 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. Thus, \(T\) is not one-to-one, according to theorem 12. Moreover, the column of \(A\) has a pivot in each row; so the rows of \(A\) spans \({\mathbb{R}^2}\). Therefore,\(T\) maps \({\mathbb{R}^3}\), according to theorem 12.

Thus, the specified linear transformation is onto.