Question

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

15. \(\left[ {\begin{array}{*{20}{c}}1\\{\bf{0}}\\{ - {\bf{3}}}\\{\bf{2}}\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{1}}\\{\bf{2}}\\{ - {\bf{3}}}\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}{ - {\bf{3}}}\\{ - {\bf{4}}}\\{\bf{1}}\\{\bf{6}}\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{3}}}\\{ - {\bf{8}}}\\7\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}{\bf{2}}\\{\bf{1}}\\{ - {\bf{6}}}\\{\bf{9}}\end{array}} \right]\)

Short answer

The basis for the space spanned by the vectors is \(\left\{ {\left[ {\begin{array}{*{20}{c}}1\\0\\{ - 3}\\2\end{array}} \right],\left[ {\begin{array}{*{20}{c}}0\\1\\2\\{ - 3}\end{array}} \right],\left[ {\begin{array}{*{20}{c}}1\\{ - 3}\\{ - 8}\\7\end{array}} \right]} \right\}\).

Step-by-step solution

State the basis for Col A

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

Obtain the row-reduced echelon form

Consider the vectors\(\left[ {\begin{array}{*{20}{c}}1\\0\\{ - 3}\\2\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}0\\1\\2\\{ - 3}\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 4}\\1\\6\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}1\\{ - 3}\\{ - 8}\\7\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}2\\1\\{ - 6}\\9\end{array}} \right]\).

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

\(A = \left[ {\begin{array}{*{20}{c}}1&0&{ - 3}&1&2\\0&1&{ - 4}&{ - 3}&1\\{ - 3}&2&1&{ - 8}&{ - 6}\\2&{ - 3}&6&7&9\end{array}} \right]\)

Obtain theechelon formof matrix A as shown below:

Add 3 times row 1 to row 3 to get row 3.

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

Add\( - 2\)times row 1 to row 4 to get row 4.

\(A = \left[ {\begin{array}{*{20}{c}}1&0&{ - 3}&1&2\\0&1&{ - 4}&{ - 3}&1\\0&2&{ - 8}&{ - 5}&0\\0&{ - 3}&{12}&5&5\end{array}} \right]\)

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

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

Add\( - 1\)timesrow 3 to row 1 to get row 1. And add 3row3 to row 2 to get row 2. Then, add 4 times row 3 to row 4 to get row 4.

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

Write the basis for Col A

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.

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

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

The corresponding columns of matrix A are shown below:

\(\left[ {\begin{array}{*{20}{c}}1\\0\\{ - 3}\\2\end{array}} \right]\),\(\left[ {\begin{array}{*{20}{c}}0\\1\\2\\{ - 3}\end{array}} \right]\),\(\left[ {\begin{array}{*{20}{c}}1\\{ - 3}\\{ - 8}\\7\end{array}} \right]\)

The column space is shown below:

\({\rm{Col }}A = \left\{ {\left[ {\begin{array}{*{20}{c}}1\\0\\{ - 3}\\2\end{array}} \right],\left[ {\begin{array}{*{20}{c}}0\\1\\2\\{ - 3}\end{array}} \right],\left[ {\begin{array}{*{20}{c}}1\\{ - 3}\\{ - 8}\\7\end{array}} \right]} \right\}\)

Thus, the basis for Col Ais \(\left\{ {\left[ {\begin{array}{*{20}{c}}1\\0\\{ - 3}\\2\end{array}} \right],\left[ {\begin{array}{*{20}{c}}0\\1\\2\\{ - 3}\end{array}} \right],\left[ {\begin{array}{*{20}{c}}1\\{ - 3}\\{ - 8}\\7\end{array}} \right]} \right\}\).