[M] According to Theorem 11, a linearly independent set \(\left\{ {{{\bf{v}}_1},....{{\bf{v}}_k}} \right\}\) in \({\mathbb{R}^n}\) can be expanded to a basis for \({\mathbb{R}^n}\). One way to this is to create \(A = \left[ {\begin{array}{*{20}{c}}{{{\bf{v}}_1}}& \cdots &{{{\bf{v}}_k}\,\;\,\begin{array}{*{20}{c}}{{{\bf{e}}_1}}& \cdots &{{{\bf{e}}_n}}\end{array}}\end{array}} \right]\), with \({{\bf{e}}_1}\),….,\({{\bf{e}}_n}\) the columns of identity matrix; the pivot columns of A form a basis for \({\mathbb{R}^n}\).

a. Use the method described to extend the following vectors to a basis for \({\mathbb{R}^5}\).

\({{\bf{v}}_1} = \left[ {\begin{array}{*{20}{c}}{ - {\bf{9}}}\\{ - {\bf{7}}}\\{\bf{8}}\\{ - {\bf{5}}}\\{\bf{7}}\end{array}} \right]\), \({{\bf{v}}_2} = \left[ {\begin{array}{*{20}{c}}{\bf{9}}\\{\bf{4}}\\{\bf{1}}\\{\bf{6}}\\{ - {\bf{7}}}\end{array}} \right]\), \({{\bf{v}}_3} = \left[ {\begin{array}{*{20}{c}}{\bf{6}}\\{\bf{7}}\\{ - {\bf{8}}}\\{\bf{5}}\\{ - {\bf{7}}}\end{array}} \right]\)

b. Explain why the method works in general: Why are the original vectors \({{\bf{v}}_1}\),….,\({{\bf{v}}_k}\) included in the basis found for Col A? Why is \({\bf{Col}}\,A = {\mathbb{R}^n}\)?

The matrix using vectors can be written as:

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

The matrix \(\left[ {\begin{array}{*{20}{c}}A&{{I_5}}\end{array}} \right]\) can be expressed as:

\(\left[ {\begin{array}{*{20}{c}}A&{{I_5}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 9}&9&6\\{ - 7}&4&7\\8&1&{ - 8}\\{ - 5}&6&5\\7&{ - 7}&{ - 7}\end{array}\,\,\,\,\,\begin{array}{*{20}{c}}1&0&0&0&0\\0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&1\end{array}} \right]\)

Let

\(B = \left[ {\begin{array}{*{20}{c}}{ - 9}&9&6\\{ - 7}&4&7\\8&1&{ - 8}\\{ - 5}&6&5\\7&{ - 7}&{ - 7}\end{array}\,\,\,\,\,\begin{array}{*{20}{c}}1&0&0&0&0\\0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&1\end{array}} \right]\).

Use the following MATLAB code to find the row reduce echelon form:

\(\begin{array}{l} > > B = \left[ { - 9\,\,9\,\,6\,\,1\,\,0\,\,0\,\,0\,\,0;\,\, - 7\,\,4\,\,7\,\,0\,\,1\,\,0\,\,0\,\,0\,;\,\,8\,\,1\,\, - 8\,\,0\,\,0\,\,1\,\,0\,\,0;\,\, - 5\,\,6\,\,5\,\,0\,\,0\,\,0\,\,1\,\,0;\,\,7\,\, - 7\,\, - 7\,\,0\,\,0\,\,0\,\,0\,\,1} \right];\\ > > U = {\rm{rref}}\left( A \right)\end{array}\)\(\left[ {\begin{array}{*{20}{c}}{ - 9}&9&6\\{ - 7}&4&7\\8&1&{ - 8}\\{ - 5}&6&5\\7&{ - 7}&{ - 7}\end{array}\,\,\,\,\,\begin{array}{*{20}{c}}1&0&0&0&0\\0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&1\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\\0&0&0\\0&0&0\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}{ - \frac{1}{3}}&0&0&1&{\frac{3}{7}}\\0&0&0&1&{\frac{5}{7}}\\{ - \frac{1}{3}}&0&0&0&{ - \frac{3}{7}}\\0&1&0&3&{\frac{{22}}{7}}\\0&0&1&0&{ - \frac{{53}}{7}}\end{array}} \right]\)

As the pivot columns are 1, 2, 3, 5, and 6, the extended basis is:

\(\left\{ {{{\bf{v}}_1},\,\,{{\bf{v}}_2},\;\;{{\bf{v}}_3},\;\;{{\bf{e}}_2},\;{{\bf{e}}_3}} \right\}\)

If the set \(\left\{ {{{\bf{v}}_1},\;{{\bf{v}}_2},\;....,\;{{\bf{v}}_k}} \right\}\) is linearly independent, there is a pivot element in each column. Therefore,

\({\rm{Col}}\left( A \right) = {\mathbb{R}^n}\)