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 the coordinate mapping is onto \({\mathbb{R}^n}\). That is, given any y in \({\mathbb{R}^n}\), with entries \({y_{\bf{1}}}\),….,\({y_n}\), produce u in V such that \({\left( {\bf{u}} \right)_B} = y\).

Short answer

The given mapping is onto.

Step-by-step solution

Find the coordinate u

\(y\)in \({\mathbb{R}^n}\) is expressed as:

\(y = \left( {\begin{array}{*{20}{c}}{{y_1}}\\{{y_2}}\\{{y_3}}\end{array}} \right)\)

The coordinates of \({\bf{u}}\) relative to basis B are the weights \({y_1}\), \({y_2}\),… \({y_n}\) such that

\({\bf{u}} = {y_1}{{\bf{b}}_1} + {y_2}{{\bf{b}}_2} + .... + {y_n}{{\bf{b}}_n}\)

Find the vector \({\left( {\bf{u}} \right)_B}\)

The coordinate of vector u is:

\(\begin{array}{c}{\left( {\bf{u}} \right)_B} = \left( {\begin{array}{*{20}{c}}{{y_1}}\\{{y_2}}\\{{y_3}}\end{array}} \right)\\ = y\end{array}\)

For any arbitrary element \(y \in {\mathbb{R}^n}\), there is an element u in V. So, every vector in \({\mathbb{R}^n}\) has a pre-image corresponding to the mapping.

Thus, the coordinate mapping is onto.