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

25. The transformation in Exercise 17.

Write the transformation in Exercise 17.

Write the transformation \(T\left( x \right)\) and \(x\) as the column vectors and fill in the entries in \(A\).

\(\begin{array}{c}T\left( x \right) = \left[ {\begin{array}{*{20}{c}}0\\{{x_1} + {x_2}}\\{{x_2} + {x_3}}\\{{x_3} + {x_4}}\end{array}} \right]\\ = \left[ A \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}0&0&0&0\\1&1&0&0\\0&1&1&0\\0&0&1&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right]\end{array}\)

Interchange row 1 and row 2.

\(\left[ {\begin{array}{*{20}{c}}1&1&0&0\\0&0&0&0\\0&1&1&0\\0&0&1&1\end{array}} \right]\)

Interchange row 2 and row 3, and row 3 and row 4.

\(\left[ {\begin{array}{*{20}{c}}1&1&0&0\\0&1&1&0\\0&0&1&1\\0&0&0&0\end{array}} \right]\)

Theorem 11states that let \(T:{\mathbb{R}^n} \to {\mathbb{R}^m}\) be a linear transformation, then \(T\) is one-to-one if and only if the equation \(T\left( x \right) = 0\) has only a trivial solution.

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 \(T\) maps \[{\mathbb{R}^n}\] onto \[{\mathbb{R}^m}\] if and only if the columns of \(A\) span \[{\mathbb{R}^m}\].

There are only three pivot positions in the matrix \(A\); so the equation \(Ax = 0\) has a nontrivial solution. The transformation \(T\) is not one-to-one, according to theorem 11. In addition, the column of \(A\) does not span \({\mathbb{R}^4}\) because \(A\) does not have a pivot in each row. \(T\) cannot map \({\mathbb{R}^4}\) onto \({\mathbb{R}^4}\), according to theorem 12.

Thus, the specified linear transformation is neither one-to-one nor onto.