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

16. \(\left[ {\begin{array}{*{20}{c}}1\\{\bf{0}}\\{\bf{0}}\\{\bf{1}}\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}{ - {\bf{2}}}\\{\bf{1}}\\{ - {\bf{1}}}\\{\bf{1}}\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}{\bf{6}}\\{ - {\bf{1}}}\\{\bf{2}}\\{ - {\bf{1}}}\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}{\bf{5}}\\{ - {\bf{3}}}\\{\bf{3}}\\{ - {\bf{4}}}\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{3}}\\{ - {\bf{1}}}\\{\bf{1}}\end{array}} \right]\)

The set of alllinear combinations of the columns of matrix A is Col A.It is called thecolumn space of A.Pivot columns are thebasisfor Col A.

Consider the vectors\(\left[ {\begin{array}{*{20}{c}}1\\0\\0\\1\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}{ - 2}\\1\\{ - 1}\\1\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}6\\{ - 1}\\2\\{ - 1}\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}5\\{ - 3}\\3\\{ - 4}\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}0\\3\\{ - 1}\\1\end{array}} \right]\).

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

\(A = \left[ {\begin{array}{*{20}{c}}1&{ - 2}&6&5&0\\0&1&{ - 1}&{ - 3}&3\\0&{ - 1}&2&3&{ - 1}\\1&1&{ - 1}&{ - 4}&1\end{array}} \right]\)

Obtain theechelon formof matrix A as shown below:

Add\( - 1\)times row 1 to row 4 to get row 4.

\(A = \left[ {\begin{array}{*{20}{c}}1&{ - 2}&6&5&0\\0&1&{ - 1}&{ - 3}&3\\0&{ - 1}&2&3&{ - 1}\\0&3&{ - 7}&{ - 9}&1\end{array}} \right]\)

Add row 2 to row 3 to get row 3.

\(A = \left[ {\begin{array}{*{20}{c}}1&{ - 2}&6&5&0\\0&1&{ - 1}&{ - 3}&3\\0&0&1&0&2\\0&3&{ - 7}&{ - 9}&1\end{array}} \right]\)

Add\( - 3\)times row 2 to row 4 to get row 4.

\(A = \left[ {\begin{array}{*{20}{c}}1&{ - 2}&6&5&0\\0&1&{ - 1}&{ - 3}&3\\0&0&1&0&2\\0&0&{ - 4}&0&{ - 8}\end{array}} \right]\)

Add 4 times row 3 to row 4 to get row 4.

\(A = \left[ {\begin{array}{*{20}{c}}1&{ - 2}&6&5&0\\0&1&{ - 1}&{ - 3}&3\\0&0&1&0&2\\0&0&0&0&0\end{array}} \right]\)

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 = \(\left[ {\begin{array}{*{20}{c}} {\boxed1}&{ - 2}&6&5&0 \\ 0&{\boxed1}&{ - 1}&{ - 3}&3 \\ 0&0&{\boxed1}&0&2 \\ 0&0&0&0&0 \end{array}} \right]\)

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

The corresponding columns of matrix A are shown below:

\(\left[ {\begin{array}{*{20}{c}}1\\0\\0\\1\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}{ - 2}\\1\\{ - 1}\\1\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}6\\{ - 1}\\2\\{ - 1}\end{array}} \right]\)

The column space is shown below:

\({\rm{Col }}A = \left\{ {\left[ {\begin{array}{*{20}{c}}1\\0\\0\\1\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 2}\\1\\{ - 1}\\1\end{array}} \right],\left[ {\begin{array}{*{20}{c}}6\\{ - 1}\\2\\{ - 1}\end{array}} \right]} \right\}\)

Thus, the basis for Col Ais \(\left\{ {\left[ {\begin{array}{*{20}{c}}1\\0\\0\\1\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 2}\\1\\{ - 1}\\1\end{array}} \right],\left[ {\begin{array}{*{20}{c}}6\\{ - 1}\\2\\{ - 1}\end{array}} \right]} \right\}\).