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.

40. \(\left[ {\begin{array}{*{20}{c}}9&{13}&5&6&{ - 1}\\{14}&{15}&{ - 7}&{ - 6}&4\\{ - 8}&{ - 9}&{12}&{ - 5}&{ - 9}\\{ - 5}&{ - 6}&{ - 8}&9&8\\{13}&{14}&{15}&2&{11}\end{array}} \right]\)

Short answer

Transformation T does not map \({\mathbb{R}^{\bf{5}}}\) onto \({\mathbb{R}^{\bf{5}}}\).

Step-by-step solution

Identify the condition for onto mapping

The transformation maps\({\mathbb{R}^n}\)onto\({\mathbb{R}^m}\)if at least one solution exists for\(T\left( {\bf{x}} \right) = {\bf{b}}\), and each vector b is in the codomain\({\mathbb{R}^m}\).

In the map \({\mathbb{R}^m} \to {\mathbb{R}^m}\), if the columns of the standard matrix span \({\mathbb{R}^m}\), there should be a pivot in every row.

Convert the matrix into the row-reduced echelon form

Consider the matrix \(A = \left[ {\begin{array}{*{20}{c}}9&{13}&5&6&{ - 1}\\{14}&{15}&{ - 7}&{ - 6}&4\\{ - 8}&{ - 9}&{12}&{ - 5}&{ - 9}\\{ - 5}&{ - 6}&{ - 8}&9&8\\{13}&{14}&{15}&2&{11}\end{array}} \right]\).

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

\(\begin{array}{l} > > {\rm{ A }} = {\rm{ }}\left[ \begin{array}{l}{\rm{9 13 5 6 }} - {\rm{1}};{\rm{ 14 15 }} - {\rm{7 }} - {\rm{6 4}};{\rm{ }} - 8{\rm{ }} - {\rm{9 12 }} - 5{\rm{ }} - {\rm{9}};\\ - {\rm{5 }} - 6{\rm{ }} - {\rm{8 9 8; 13 14 15 2 11}}\end{array} \right];\\ > > {\rm{ U}} = {\rm{rref}}\left( {\rm{A}} \right)\end{array}\)

\(\left[ {\begin{array}{*{20}{c}}9&{13}&5&6&{ - 1}\\{14}&{15}&{ - 7}&{ - 6}&4\\{ - 8}&{ - 9}&{12}&{ - 5}&{ - 9}\\{ - 5}&{ - 6}&{ - 8}&9&8\\{13}&{14}&{15}&2&{11}\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&0&0&0&5\\0&1&0&0&{ - 4}\\0&0&1&0&0\\0&0&0&1&1\\0&0&0&0&0\end{array}} \right]\)

In the obtained matrix, the fifth column does not have a pivot position. So, it does not span \({\mathbb{R}^5}\).

Thus, transformation T does not map \({\mathbb{R}^{\bf{5}}}\) onto \({\mathbb{R}^{\bf{5}}}\).