Question:In Exercises 15–18, find a basis for the space spanned by the given vectors,\({{\bf{v}}_{\bf{1}}}, \ldots ,{{\bf{v}}_{\bf{5}}}\).

18. \(\left( {\begin{array}{*{20}{c}}{ - 8}\\{\bf{7}}\\{\bf{6}}\\{\bf{5}}\\{ - {\bf{7}}}\end{array}} \right)\), \(\left( {\begin{array}{*{20}{c}}{\bf{8}}\\{ - {\bf{7}}}\\{ - {\bf{9}}}\\{ - {\bf{5}}}\\{\bf{7}}\end{array}} \right)\), \(\left( {\begin{array}{*{20}{c}}{ - {\bf{8}}}\\{\bf{7}}\\{\bf{4}}\\{\bf{5}}\\{ - {\bf{7}}}\end{array}} \right)\), \(\left( {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{4}}\\{\bf{9}}\\{\bf{6}}\\{ - {\bf{7}}}\end{array}} \right)\), \(\left( {\begin{array}{*{20}{c}}{ - {\bf{9}}}\\{\bf{3}}\\{ - {\bf{4}}}\\{ - {\bf{1}}}\\{\bf{0}}\end{array}} \right)\)

The set of alllinear combinations of the columns of matrix A is Col A.It is called thecolumn space of A.Pivotcolumns are thebasis for Col A.

Consider the vectors \(\left( {\begin{array}{*{20}{c}}{ - 8}\\7\\6\\5\\{ - 7}\end{array}} \right)\), \(\left( {\begin{array}{*{20}{c}}8\\{ - 7}\\{ - 9}\\{ - 5}\\7\end{array}} \right)\), \(\left( {\begin{array}{*{20}{c}}{ - 8}\\7\\4\\5\\{ - 7}\end{array}} \right)\), \(\left( {\begin{array}{*{20}{c}}1\\4\\9\\6\\{ - 7}\end{array}} \right)\), \(\left( {\begin{array}{*{20}{c}}{ - 9}\\3\\{ - 4}\\{ - 1}\\0\end{array}} \right)\).

The five vectors span the column spaceof a matrix. So, construct matrix A by using the given vectors as shown below:

\(A = \left( {\begin{array}{*{20}{c}}{ - 8}&8&{ - 8}&1&{ - 9}\\7&{ - 7}&7&4&3\\6&{ - 9}&4&9&{ - 4}\\5&{ - 5}&5&6&{ - 1}\\{ - 7}&7&{ - 7}&{ - 7}&0\end{array}} \right)\)

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

\( > > {\rm{ U}} = {\rm{rref}}\left( {\rm{A}} \right)\)

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

To identify the pivot and the pivot position, observe the leftmost column (nonzero column) of the matrix, that is, the pivot column. At the top of this column, 1 is the pivot.

\(A = \left[ {\begin{array}{*{20}{c}} {\boxed1}&0&{\frac{5}{3}}&0&{\frac{4}{3}} \\ 0&{\boxed1}&{\frac{2}{3}}&0&{\frac{1}{3}} \\ 0&0&0&{\boxed1}&{ - 1} \\ 0&0&0&0&0 \\ 0&0&0&0&0 \end{array}} \right]\)

The first, second, and fourth columns have pivot elements.

The corresponding columns of matrix A are shown below:

\(\left( {\begin{array}{*{20}{c}}{ - 8}\\7\\6\\5\\{ - 7}\end{array}} \right)\),\(\left( {\begin{array}{*{20}{c}}8\\{ - 7}\\{ - 9}\\{ - 5}\\7\end{array}} \right)\),\(\left( {\begin{array}{*{20}{c}}1\\4\\9\\6\\{ - 7}\end{array}} \right)\)

The column space is shown below:

\({\rm{Col }}A = \left\{ {\left( {\begin{array}{*{20}{c}}{ - 8}\\7\\6\\5\\{ - 7}\end{array}} \right),\left( {\begin{array}{*{20}{c}}8\\{ - 7}\\{ - 9}\\{ - 5}\\7\end{array}} \right),\left( {\begin{array}{*{20}{c}}1\\4\\9\\6\\{ - 7}\end{array}} \right)} \right\}\)

Thus, the basis for Col Ais \(\left\{ {\left( {\begin{array}{*{20}{c}}{ - 8}\\7\\6\\5\\{ - 7}\end{array}} \right),\left( {\begin{array}{*{20}{c}}8\\{ - 7}\\{ - 9}\\{ - 5}\\7\end{array}} \right),\left( {\begin{array}{*{20}{c}}1\\4\\9\\6\\{ - 7}\end{array}} \right)} \right\}\).