Question

Exercises 23-26 concern a vector space V, a basis \(B = \left\{ {{{\bf{b}}_{\bf{1}}},....,{{\bf{b}}_n}\,} \right\}\) and the coordinate mapping \({\bf{x}} \mapsto {\left( {\bf{x}} \right)_B}\).

Show that a subset \(\left\{ {{{\bf{u}}_1},...,{{\bf{u}}_p}} \right\}\) in V is linearly independent if and only if the set of coordinate vectors \(\left\{ {{{\left( {{{\bf{u}}_{\bf{1}}}} \right)}_B},.....,{{\left( {{{\bf{u}}_p}} \right)}_B}} \right\}\) is linearly independent in \({\mathbb{R}^n}\)(Hint: Since the coordinate mapping is one-to-one, the following equations have the same solutions, \({c_{\bf{1}}}\),….,\({c_p}\))

\({c_{\bf{1}}}{{\bf{u}}_{\bf{1}}} + ..... + {c_p}{{\bf{u}}_p} = {\bf{0}}\) The zero vector V

\({\left( {{c_{\bf{1}}}{{\bf{u}}_{\bf{1}}} + ..... + {c_p}{{\bf{u}}_p}} \right)_B} = {\left( {\bf{0}} \right)_B}\) The zero vector in \({\mathbb{R}^n}\)a

Short answer

The given statement is true.

Step-by-step solution

Check the solutions for the subset in V and coordinate vectors

If the coordinate mappingis linear, the equations \({c_1}{{\bf{u}}_1} + ... + {c_p}{{\bf{u}}_p} = 0\) and \(\left( {{c_1}{{\bf{u}}_1} + ..... + {c_p}{{\bf{u}}_p}} \right] = {\left( 0 \right]_B}\) should have the same solutions.

The first equation will have a trivial solutionif the second equation also has a trivial solution.

Check for linear independence

According to the definition of linear independence, the set \(\left\{ {{{\bf{u}}_1},....,{{\bf{u}}_p}} \right\}\) is linearly independent if \(\left\{ {{{\left( {{{\bf{u}}_1}} \right]}_B},.....,{{\left( {{{\bf{u}}_p}} \right]}_B}} \right\}\) is linearly independent.

Thus, the given statement is true.