91Ó°ÊÓ

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.

Consider the vectors\({{\bf{v}}_1} = \left( {\begin{array}{*{20}{c}}7\\4\\{ - 9}\\{ - 5}\end{array}} \right)\), \({{\bf{v}}_2} = \left( {\begin{array}{*{20}{c}}4\\{ - 7}\\2\\5\end{array}} \right)\), \({{\bf{v}}_3} = \left( {\begin{array}{*{20}{c}}1\\{ - 5}\\3\\4\end{array}} \right)\).

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

\(\begin{array}{c}\left( {\begin{array}{*{20}{c}}7\\4\\{ - 9}\\{ - 5}\end{array}} \right) - 3\left( {\begin{array}{*{20}{c}}4\\{ - 7}\\2\\5\end{array}} \right) + 5\left( {\begin{array}{*{20}{c}}1\\{ - 5}\\3\\4\end{array}} \right) = 0\\\left( {\begin{array}{*{20}{c}}7\\4\\{ - 9}\\{ - 5}\end{array}} \right) - \left( {\begin{array}{*{20}{c}}{12}\\{ - 21}\\6\\{15}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}5\\{ - 25}\\{15}\\{20}\end{array}} \right) = 0\\\left( {\begin{array}{*{20}{c}}{7 - 12 + 5}\\{4 + 21 - 25}\\{ - 9 - 6 + 15}\\{ - 5 - 15 + 20}\end{array}} \right) = 0\\0 = 0\end{array}\)

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

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

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

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

\({{\bf{v}}_2} = \frac{1}{3}{{\bf{v}}_1} + \frac{5}{3}{{\bf{v}}_3}\)

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

\({{\bf{v}}_3} = - \frac{1}{5}{{\bf{v}}_1} + \frac{3}{5}{{\bf{v}}_2}\)

Thus, all vectors are linear combinations of each other.

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

From the equation\({{\bf{v}}_2} = \frac{1}{3}{{\bf{v}}_1} + \frac{5}{3}{{\bf{v}}_3}\), the set\(\left\{ {{{\bf{v}}_1},{{\bf{v}}_3}} \right\}\)spans H.

From the equation\({{\bf{v}}_3} = - \frac{1}{5}{{\bf{v}}_1} + \frac{3}{5}{{\bf{v}}_2}\), the set\(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2}} \right\}\)spans 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.