Question

Question: 13. Suppose \(\left\{ {{{\rm{v}}_{\rm{1}}}{\rm{,}}{{\rm{v}}_{\rm{2}}}{\rm{,}}{{\rm{v}}_{\rm{3}}}} \right\}\) is a basis for \({\mathbb{R}^3}\). Show that Span \(\left\{ {{{\rm{v}}_{\rm{2}}} - {{\rm{v}}_{\rm{1}}},{{\rm{v}}_{\rm{3}}} - {{\rm{v}}_{\rm{1}}}} \right\}\) is a plane in \({\mathbb{R}^3}\). (Hint: What can you say about \({\rm{u}}\) and \({\rm{v}}\)when Span \(\left\{ {{\rm{u,v}}} \right\}\) is a plane?)

Short answer

It is shown that the Span of \(\left\{ {{{\rm{v}}_{\rm{2}}} - {{\rm{v}}_{\rm{1}}},{{\rm{v}}_{\rm{3}}} - {{\rm{v}}_{\rm{1}}}} \right\}\)is in the plane \({\mathbb{R}^3}\).

Step-by-step solution

Describe the given statement

It is given that \(\left\{ {{{\rm{v}}_{\rm{1}}}{\rm{,}}{{\rm{v}}_{\rm{2}}}{\rm{,}}{{\rm{v}}_{\rm{3}}}} \right\}\) is a basis for \({\mathbb{R}^3}\)and it is to be shown that Span of \(\left\{ {{{\rm{v}}_{\rm{2}}} - {{\rm{v}}_{\rm{1}}},{{\rm{v}}_{\rm{3}}} - {{\rm{v}}_{\rm{1}}}} \right\}\)is in the plane \({\mathbb{R}^3}\). This is possible only if the set of vectors \(\left\{ {{{\rm{v}}_{\rm{2}}} - {{\rm{v}}_{\rm{1}}},{{\rm{v}}_{\rm{3}}} - {{\rm{v}}_{\rm{1}}}} \right\}\)is linearly independent.

 Make a linear combination of the given set of vectors

A possible linear combination of a set of vectors \(\left\{ {{{\rm{v}}_{\rm{2}}} - {{\rm{v}}_{\rm{1}}},{{\rm{v}}_{\rm{3}}} - {{\rm{v}}_{\rm{1}}}} \right\}\)is \({c_1}\left( {{{\bf{v}}_2} - {{\bf{v}}_1}} \right) + {c_2}\left( {{{\bf{v}}_3} - {{\bf{v}}_1}} \right)\), where \({c_2},{c_3}\) satisfies \({c_1}\left( {{{\rm{v}}_{\rm{2}}} - {{\rm{v}}_{\rm{1}}}} \right) + {c_2}\left( {{{\rm{v}}_{\rm{3}}} - {{\rm{v}}_{\rm{1}}}} \right) = 0\). If \({c_1}\left( {{{\rm{v}}_{\rm{2}}} - {{\rm{v}}_{\rm{1}}}} \right) + {c_2}\left( {{{\rm{v}}_{\rm{3}}} - {{\rm{v}}_{\rm{1}}}} \right) = 0\) is linearly independent, then \({c_1}\left( {{{\rm{v}}_{\rm{2}}} - {{\rm{v}}_{\rm{1}}}} \right) + {c_2}\left( {{{\rm{v}}_{\rm{3}}} - {{\rm{v}}_{\rm{1}}}} \right) = 0\) must be equal to 0, which is possible only if \({c_1} = {c_2} = 0\).

 Use property of a linear independent set

According to the property of linear independent set, if \(\left\{ {{{\rm{v}}_{\rm{1}}}{\rm{,}}{{\rm{v}}_{\rm{2}}}{\rm{,}}{{\rm{v}}_{\rm{3}}}} \right\}\) is a set of linearly independent vectors, then all the coefficients of its linear combination are 0. So, \({c_1} = {c_2} = 0\) is true for all \(\left\{ {{{\rm{v}}_{\rm{1}}}{\rm{,}}{{\rm{v}}_{\rm{2}}}{\rm{,}}{{\rm{v}}_{\rm{3}}}} \right\}\) in \({\mathbb{R}^3}\).

 Draw a conclusion

As the set of vectors \(\left\{ {{{\rm{v}}_{\rm{2}}} - {{\rm{v}}_{\rm{1}}},{{\rm{v}}_{\rm{3}}} - {{\rm{v}}_{\rm{1}}}} \right\}\)is linearly independent, so the Span of \(\left\{ {{{\rm{v}}_{\rm{2}}} - {{\rm{v}}_{\rm{1}}},{{\rm{v}}_{\rm{3}}} - {{\rm{v}}_{\rm{1}}}} \right\}\)is in the plane \({\mathbb{R}^3}\).

Hence, it is proved that the Span of \(\left\{ {{{\rm{v}}_{\rm{2}}} - {{\rm{v}}_{\rm{1}}},{{\rm{v}}_{\rm{3}}} - {{\rm{v}}_{\rm{1}}}} \right\}\)is in the plane \({\mathbb{R}^3}\).