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 \({\bf{x}} \mapsto {\left( {\bf{x}} \right)_B}\).

Given vectors, \({u_{\bf{1}}}\),….,\({u_p}\) and w in V, show that w is a linear combination of \({u_{\bf{1}}}\),….,\({u_p}\) if and only if \({\left( w \right)_B}\) is a linear combination of vectors \({\left( {{{\bf{u}}_{\bf{1}}}} \right)_B}\),….,\({\left( {{{\bf{u}}_p}} \right)_B}\).

Short answer

\({\left( {\bf{w}} \right)_B}\) can be written as a linear combination of \({\left( {{{\bf{u}}_1}} \right)_B}\), \({\left( {{{\bf{u}}_2}} \right)_B}\),….,\({\left( {{{\bf{u}}_p}} \right)_B}\).

Step-by-step solution

Write w as a linear combination

Was a linear combinationof \({{\bf{u}}_1}\), \({{\bf{u}}_2}\),…., \({{\bf{u}}_p}\) can be written as:

\({\bf{w}} = {c_1}{{\bf{u}}_1} + {c_2}{{\bf{u}}_2} + .... + {c_p}{{\bf{u}}_p}\)

The first equation has a trivial solution if the second equation also has a trivial solution.

Check for \({\left( {\bf{w}} \right)_B}\)

If B is a basis for space V, then the coordinate mapping is one to one transformation. Therefore,

\(\begin{array}{c}{\left( {\bf{w}} \right)_B} = {\left( {{c_1}{{\bf{u}}_1} + {c_2}{{\bf{u}}_2} + .....{c_p}{{\bf{u}}_p}} \right)_B}\\ = {\left( {{c_1}{{\bf{u}}_1}} \right)_B} + {\left( {{c_2}{{\bf{u}}_2}} \right)_B} + .... + {\left( {{c_p}{{\bf{u}}_p}} \right)_B}\\ = {c_1}{\left( {{{\bf{u}}_1}} \right)_B} + {c_2}{\left( {{{\bf{u}}_2}} \right)_B} + .... + {c_p}{\left( {{{\bf{u}}_p}} \right)_B}\end{array}\)

Thus, the \({\left( {\bf{w}} \right)_B}\) can be written as a linear combination of \({\left( {{{\bf{u}}_1}} \right)_B}\), \({\left( {{{\bf{u}}_2}} \right)_B}\),….,\({\left( {{{\bf{u}}_p}} \right)_B}\).