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

17. \(\left( {\begin{array}{*{20}{c}}8\\9\\{ - 3}\\{ - 6}\\0\end{array}} \right)\), \(\left( {\begin{array}{*{20}{c}}{\bf{4}}\\{\bf{5}}\\{\bf{1}}\\{ - {\bf{4}}}\\{\bf{4}}\end{array}} \right)\), \(\left( {\begin{array}{*{20}{c}}{ - {\bf{1}}}\\{ - {\bf{4}}}\\{ - {\bf{9}}}\\{\bf{6}}\\{ - {\bf{7}}}\end{array}} \right)\), \(\left( {\begin{array}{*{20}{c}}{\bf{6}}\\{\bf{8}}\\{\bf{4}}\\{ - {\bf{7}}}\\{{\bf{10}}}\end{array}} \right)\), \(\left( {\begin{array}{*{20}{c}}{ - {\bf{1}}}\\{\bf{4}}\\{{\bf{11}}}\\{ - {\bf{8}}}\\{ - {\bf{7}}}\end{array}} \right)\)

The set of all linear combinations of the columns of matrix A is Col A.It is called the column space of A. Pivotcolumns are the basis for Col A.

Consider the vectors\(\left( {\begin{array}{*{20}{c}}8\\9\\{ - 3}\\{ - 6}\\0\end{array}} \right)\),\(\left( {\begin{array}{*{20}{c}}4\\5\\1\\{ - 4}\\4\end{array}} \right)\),\(\left( {\begin{array}{*{20}{c}}{ - 1}\\{ - 4}\\{ - 9}\\6\\{ - 7}\end{array}} \right)\),\(\left( {\begin{array}{*{20}{c}}6\\8\\4\\{ - 7}\\{10}\end{array}} \right)\),\(\left( {\begin{array}{*{20}{c}}{ - 1}\\4\\{11}\\{ - 8}\\{ - 7}\end{array}} \right)\).

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

\(A = \left( {\begin{array}{*{20}{c}}8&4&{ - 1}&6&{ - 1}\\9&5&{ - 4}&8&4\\{ - 3}&1&{ - 9}&4&{11}\\{ - 6}&{ - 4}&6&{ - 7}&{ - 8}\\0&4&{ - 7}&{10}&{ - 7}\end{array}} \right)\)

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

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

\(\left( {\begin{array}{*{20}{c}}8&4&{ - 1}&6&{ - 1}\\9&5&{ - 4}&8&4\\{ - 3}&1&{ - 9}&4&{11}\\{ - 6}&{ - 4}&6&{ - 7}&{ - 8}\\0&4&{ - 7}&{10}&{ - 7}\end{array}} \right) \sim \left( {\begin{array}{*{20}{c}}1&0&0&{ - 1/2}&3\\0&1&0&{5/2}&{ - 7}\\0&0&1&0&{ - 3}\\0&0&0&0&0\\0&0&0&0&0\end{array}} \right)\)

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

\(A = \left[ {\begin{array}{*{20}{c}} {\boxed1}&0&0&{ - \frac{1}{2}}&3 \\ 0&{\boxed1}&0&{\frac{5}{2}}&{ - 7} \\ 0&0&{\boxed1}&0&{ - 3} \\ 0&0&0&0&0 \\ 0&0&0&0&0 \end{array}} \right]\)

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

The corresponding columns of matrix A are shown below:

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

The column space is shown below:

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

Thus, the basis for Col Ais \(\left\{ {\left( {\begin{array}{*{20}{c}}8\\9\\{ - 3}\\{ - 6}\\0\end{array}} \right),\left( {\begin{array}{*{20}{c}}4\\5\\1\\{ - 4}\\4\end{array}} \right),\left( {\begin{array}{*{20}{c}}{ - 1}\\{ - 4}\\{ - 9}\\6\\{ - 7}\end{array}} \right)} \right\}\).