Question

In Exercises 37–40, let T be the linear transformation whose standard matrix is given. In Exercises 37 and 38, decide if T is a one-to-one mapping. In Exercises 39 and 40, decide if T maps \({\mathbb{R}^{\bf{5}}}\) onto \({\mathbb{R}^{\bf{5}}}\). Justify your answers.

37. \(\left[ {\begin{array}{*{20}{c}}{ - 5}&{10}&{ - 5}&4\\8&3&{ - 4}&7\\4&{ - 9}&5&{ - 3}\\{ - 3}&{ - 2}&5&4\end{array}} \right]\)

Short answer

Transformation T is not one-to-one.

Step-by-step solution

Identify the condition for one-to-one mapping

The transformation maps\({\mathbb{R}^n}\)one-to-one\({\mathbb{R}^m}\)if at most one solution exists for\(T\left( {\bf{x}} \right) = {\bf{b}}\), and each vector b is in the codomain \({\mathbb{R}^m}\). And this linear transformation, \(T:{\mathbb{R}^n} \to {\mathbb{R}^m}\), has only a trivial solution.

Convert the matrix into the row-reduced echelon form

Consider the matrix\(A = \left[ {\begin{array}{*{20}{c}}{ - 5}&{10}&{ - 5}&4\\8&3&{ - 4}&7\\4&{ - 9}&5&{ - 3}\\{ - 3}&{ - 2}&5&4\end{array}} \right]\).

Use the code in the MATLAB to obtain the row-reduced echelon form, as shown below:

\(\begin{array}{l} > > {\rm{ A }} = {\rm{ }}\left[ { - 5{\rm{ }}10{\rm{ }} - 5{\rm{ }}4;{\rm{ }}8{\rm{ }}3{\rm{ }} - 4{\rm{ }}7;{\rm{ }}4{\rm{ }} - 9{\rm{ }}5{\rm{ }} - 3;{\rm{ }} - 3{\rm{ }} - 2{\rm{ }}5{\rm{ }}4} \right];\\ > > {\rm{ U}} = {\rm{rref}}\left( {\rm{A}} \right)\end{array}\)

\(\left[ {\begin{array}{*{20}{c}}{ - 5}&{10}&{ - 5}&4\\8&3&{ - 4}&7\\4&{ - 9}&5&{ - 3}\\{ - 3}&{ - 2}&5&4\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&0&0&{44/35}\\0&1&0&{79/35}\\0&0&1&{86/35}\\0&0&0&0\end{array}} \right]\)

You can also write it as \(\left[ {\begin{array}{*{20}{c}}1&0&0&{1.2571}\\0&1&0&{2.2571}\\0&0&1&{2.4571}\\0&0&0&0\end{array}} \right]\).

In the obtained matrix, the fourth column does not have a pivot position. So, the solution is non-trivial.

Thus, transformation T is not one-to-one.