Question

Show that Bis a\(H = Span\left\{ {{v_1},{v_2},{v_3}} \right\}\)basis for H and x is in H, and find the B-coordinate vector of x, for

\({{\bf{v}}_1} = \left( {\begin{array}{*{20}{c}}{ - 6}\\4\\{ - 9}\\4\end{array}} \right)\),\({{\bf{v}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{8}}\\{ - {\bf{3}}}\\{\bf{7}}\\{ - {\bf{3}}}\end{array}} \right)\),\({{\bf{v}}_{\bf{3}}} = \left( {\begin{array}{*{20}{c}}{ - {\bf{9}}}\\{\bf{5}}\\{ - {\bf{8}}}\\{\bf{3}}\end{array}} \right)\),\({\bf{x}} = \left( {\begin{array}{*{20}{c}}{\bf{4}}\\{\bf{7}}\\{ - {\bf{8}}}\\{\bf{3}}\end{array}} \right)\)

Short answer

It shows thatB is a basis for H, and x is in H.

TheB-coordinate vector is \({\left( {\bf{x}} \right)_B} = \left( {\begin{array}{*{20}{c}}3\\5\\2\end{array}} \right)\).

Step-by-step solution

State the condition for the basis

If thevectorequation is represented as \({x_1}{{\bf{v}}_1} + {x_2}{{\bf{v}}_2} + \ldots + {x_p}{{\bf{v}}_p} = {\bf{x}}\), and it has a solution, then vector x is in\(H = {\rm{Span}}\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},...,{{\bf{v}}_p}} \right\}\), and it is said to be abasis for H.

Show that B is a basis for H

Consider the vectors\({{\bf{v}}_1} = \left( {\begin{array}{*{20}{c}}{ - 6}\\4\\{ - 9}\\4\end{array}} \right)\),\({{\bf{v}}_2} = \left( {\begin{array}{*{20}{c}}8\\{ - 3}\\7\\{ - 3}\end{array}} \right)\),\({{\bf{v}}_3} = \left( {\begin{array}{*{20}{c}}{ - 9}\\5\\{ - 8}\\3\end{array}} \right)\),\({\bf{x}} = \left( {\begin{array}{*{20}{c}}4\\7\\{ - 8}\\3\end{array}} \right)\).

Write the augmented matrix\(\left( {\begin{array}{*{20}{c}}{{{\bf{v}}_1}}&{{{\bf{v}}_2}}&{{{\bf{v}}_3}}&{\bf{x}}\end{array}} \right)\) as shown below:

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

Use the code in MATLAB to obtain the row-reducedechelon formas shown below:

\(\begin{array}{l} > > {\rm{ A }} = {\rm{ }}\left( { - {\rm{6 8 }} - {\rm{9 4; 4 }} - {\rm{3 5 7; }} - {\rm{9 7 }} - {\rm{8 }} - {\rm{8; 4 }} - {\rm{3 3 3}}} \right);\\ > > {\rm{ U}} = {\rm{rref}}\left( {\rm{A}} \right)\end{array}\)

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

The first, second, and third columns have pivots. So, B is a basis for H.

As the system of equations has a solution, x is in H.

Compute the B-coordinate of vector x

From the above matrix form, the solution is\({x_1} = 3\),\({x_2} = 5\), and\({x_3} = 2\).

Obtain the B-coordinate vector for x as shown below:

\(\begin{array}{c}{x_1}{{\bf{v}}_1} + {x_2}{{\bf{v}}_2} + {x_3}{{\bf{v}}_3} = {\bf{x}}\\3{{\bf{v}}_1} + 5{{\bf{v}}_2} + 2{{\bf{v}}_3} = {\bf{x}}\end{array}\)

Thus, the B-coordinate vector is \({\left( {\bf{x}} \right)_B} = \left( {\begin{array}{*{20}{c}}3\\5\\2\end{array}} \right)\).