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 \({\mathbb{R}^{\bf{5}}}\)mapping. In Exercises 39 and 40, decide if T maps onto \({\mathbb{R}^{\bf{5}}}\). Justify your answers.

39. \(\left[ {\begin{array}{*{20}{c}}4&{ - 7}&3&7&5\\6&{ - 8}&5&{12}&{ - 8}\\{ - 7}&{10}&{ - 8}&{ - 9}&{14}\\3&{ - 5}&4&2&{ - 6}\\{ - 5}&6&{ - 6}&{ - 7}&3\end{array}} \right]\)

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\(A = \left[ {\begin{array}{*{20}{c}}4&{ - 7}&3&7&5\\6&{ - 8}&5&{12}&{ - 8}\\{ - 7}&{10}&{ - 8}&{ - 9}&{14}\\3&{ - 5}&4&2&{ - 6}\\{ - 5}&6&{ - 6}&{ - 7}&3\end{array}} \right]\).

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

\(\begin{array}{l} > > {\rm{ A }} = {\rm{ }}\left[ \begin{array}{l}{\rm{4 }} - {\rm{7 3 7 5}};{\rm{ 6 }} - {\rm{8 5 12 }} - {\rm{8}};{\rm{ }} - {\rm{7 10 }} - {\rm{8 }} - {\rm{9 14}};\\{\rm{ 3 }} - 5{\rm{ 4 2 }} - {\rm{6; }} - {\rm{5 6 }} - {\rm{6 }} - {\rm{7 3}}\end{array} \right];\\ > > {\rm{ U}} = {\rm{rref}}\left( {\rm{A}} \right)\end{array}\)

\(\left[ {\begin{array}{*{20}{c}}4&{ - 7}&3&7&5\\6&{ - 8}&5&{12}&{ - 8}\\{ - 7}&{10}&{ - 8}&{ - 9}&{14}\\3&{ - 5}&4&2&{ - 6}\\{ - 5}&6&{ - 6}&{ - 7}&3\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&0&0&5&0\\0&1&0&1&0\\0&0&1&{ - 2}&0\\0&0&0&0&1\\0&0&0&0&0\end{array}} \right]\)

In the obtained matrix, the fourth 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 \({\mathbb{R}^{\bf{5}}}\).