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 the coordinate mapping is one to one. (Hint: Suppose \({\left( {\bf{u}} \right)_B} = {\left( {\bf{w}} \right)_B}\) for some u and w in V, and show that \({\bf{u}} = {\bf{w}}\)).

The basis for uand v is V.

\(\begin{array}{l}{\bf{u}} = {p_1}{{\bf{b}}_1} + .... + {p_n}{{\bf{b}}_n}\\{\bf{w}} = {q_1}{{\bf{b}}_1} + .... + {q_n}{{\bf{b}}_n}\end{array}\)

So, the changes of coordinate matrices are written as:

\({\left( {\bf{u}} \right)_B} = \left( {\begin{array}{*{20}{c}}{{p_1}}\\ \vdots \\{{p_n}}\end{array}} \right)\) and \({\left( {\bf{w}} \right)_B} = \left( {\begin{array}{*{20}{c}}{{q_1}}\\ \vdots \\{{q_n}}\end{array}} \right)\)

If \({\left( {\bf{u}} \right)_B} = {\left( {\bf{w}} \right)_B}\), then

\(\left( {\begin{array}{*{20}{c}}{{p_1}}\\ \vdots \\{{p_n}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{{q_1}}\\ \vdots \\{{q_n}}\end{array}} \right)\)

\({p_1}{{\bf{b}}_1} = {q_1}{{\bf{b}}_1},\;{p_2}{{\bf{b}}_2} = {q_2}{{\bf{b}}_2},....,{p_n}{{\bf{b}}_n} = {q_n}{{\bf{b}}_n}\).

So, \({\bf{u}} = {\bf{w}}\)

Thus, the given mapping is one to one.