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Chapter 4: Q19E (page 191)

Let \({v_1} = \left( {\begin{array}{*{20}{c}}4\\{ - 3}\\7\end{array}} \right)\), \({v_2} = \left( {\begin{array}{*{20}{c}}1\\9\\{ - 2}\end{array}} \right)\), \({v_3} = \left( {\begin{array}{*{20}{c}}7\\{11}\\6\end{array}} \right)\), and \(H = Span\left\{ {{v_1},{v_2},{v_3}} \right\}\). It can be verified that \(4{v_1} + 5{v_2} - 3{v_3} = 0\). Use this information to find a basis for \(H\). There is more than one answer

Short Answer

Expert verified

Each vector forms a basis for H.

Step by step solution

01

State the condition for span and linear independence

Vector \({{\bf{v}}_3}\) is said to be in the Span \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2}} \right\}\) if the equation \({x_1}{{\bf{v}}_1} + {x_2}{{\bf{v}}_2} = {{\bf{v}}_3}\) has a solution, where \({{\bf{v}}_1}\), \({{\bf{v}}_2}\), and \({{\bf{v}}_3}\) are vectors.

The vectors are said to be linearly independent if the equation \({x_1}{{\bf{v}}_1} + {x_2}{{\bf{v}}_2} + {x_3}{{\bf{v}}_3} + ... + {x_p}{{\bf{v}}_p} = 0\) has a trivial solution, where \({{\bf{v}}_1},{{\bf{v}}_2},...,{{\bf{v}}_p}\) are vectors.

02

Verify the condition

Consider the vectors\({{\bf{v}}_1} = \left( {\begin{array}{*{20}{c}}4\\{ - 3}\\7\end{array}} \right)\), \({{\bf{v}}_2} = \left( {\begin{array}{*{20}{c}}1\\9\\{ - 2}\end{array}} \right)\), \({{\bf{v}}_3} = \left( {\begin{array}{*{20}{c}}7\\{11}\\6\end{array}} \right)\).

Substitute the above vectors in the equation\(4{{\bf{v}}_1} + 5{{\bf{v}}_2} - 3{{\bf{v}}_3} = 0\).

\(\begin{array}{c}4\left( {\begin{array}{*{20}{c}}4\\{ - 3}\\7\end{array}} \right) + 5\left( {\begin{array}{*{20}{c}}1\\9\\{ - 2}\end{array}} \right) - 3\left( {\begin{array}{*{20}{c}}7\\{11}\\6\end{array}} \right) = 0\\\left( {\begin{array}{*{20}{c}}{16}\\{ - 12}\\{28}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}5\\{45}\\{ - 10}\end{array}} \right) - \left( {\begin{array}{*{20}{c}}{21}\\{33}\\{18}\end{array}} \right) = 0\\\left( {\begin{array}{*{20}{c}}{15 + 5 - 21}\\{ - 12 + 45 - 33}\\{28 - 10 - 18}\end{array}} \right) = 0\\0 = 0\end{array}\)

Hence, the relation\(4{{\bf{v}}_1} + 5{{\bf{v}}_2} - 3{{\bf{v}}_3} = 0\)is verified.

03

Rearrange the vectors

Take the equation\(4{{\bf{v}}_1} + 5{{\bf{v}}_2} - 3{{\bf{v}}_3} = 0\), and write vector\({{\bf{v}}_1}\)in terms of the other two vectors,\({{\bf{v}}_2}\)and\({{\bf{v}}_3}\), as shown below:

\({{\bf{v}}_1} = - \frac{5}{4}{{\bf{v}}_2} + \frac{3}{4}{{\bf{v}}_3}\)

Take the equation\(4{{\bf{v}}_1} + 5{{\bf{v}}_2} - 3{{\bf{v}}_3} = 0\), and write vector\({{\bf{v}}_2}\)in terms of the other two vectors,\({{\bf{v}}_1}\)and\({{\bf{v}}_3}\), as shown below:

\({{\bf{v}}_2} = - \frac{4}{5}{{\bf{v}}_1} + \frac{3}{5}{{\bf{v}}_3}\)

Take the equation\(4{{\bf{v}}_1} + 5{{\bf{v}}_2} - 3{{\bf{v}}_3} = 0\), and write vector\({{\bf{v}}_3}\)in terms of the other two vectors,\({{\bf{v}}_1}\)and\({{\bf{v}}_2}\), as shown below:

\({{\bf{v}}_3} = \frac{4}{3}{{\bf{v}}_1} + \frac{5}{3}{{\bf{v}}_2}\)

Thus, all vectors are linear combinations of each other.

04

Obtain the basis for H

From the equation\({{\bf{v}}_1} = - \frac{5}{4}{{\bf{v}}_2} + \frac{3}{4}{{\bf{v}}_3}\), the set\(\left\{ {{{\bf{v}}_2},{{\bf{v}}_3}} \right\}\)span H.

From the equation\({{\bf{v}}_2} = - \frac{4}{5}{{\bf{v}}_1} + \frac{3}{5}{{\bf{v}}_3}\), the set\(\left\{ {{{\bf{v}}_1},{{\bf{v}}_3}} \right\}\)span H.

From the equation\({{\bf{v}}_3} = \frac{4}{3}{{\bf{v}}_1} + \frac{5}{3}{{\bf{v}}_2}\), the set\(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2}} \right\}\)span H.

Also, no vector is a multiple of another vector.

Thus, the sets\(\left\{ {{{\bf{v}}_2},{{\bf{v}}_3}} \right\}\),\(\left\{ {{{\bf{v}}_1},{{\bf{v}}_3}} \right\}\), and\(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2}} \right\}\)are linearly independent and they all form a basis for H.

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