Exercises 15 and 16 provide a proof of Theorem 15. Fill in a justification for each step. 15.

Given v inV , there exist scalars , such that because . Apply the coordinate mapping determined by the basis , and obtain because \({\rm{v}}\)\((b)\underline {\,\,\,\,\,\,\,\,\,\,\,\,} \) . This equation may be written in the form \({\left( v \right)_C} = \left( {{{\left( {{b_1}} \right)}_C}\,\,\,{{\left( {{b_2}} \right)}_C}\,\,\,...\,\,\,\,{{\left( {{b_n}} \right)}_C}} \right)\left( \begin{array}{l}{x_1}\\\,\, \vdots \\{x_n}\end{array} \right)\)by the definition of \((c)\underline {\,\,\,\,\,\,\,\,\,\,\,\,} \) . This shows that the matrix \(\mathop P\limits_{C \to B} \)shown in (5) satisfies \({\left( v \right)_C} = \mathop P\limits_{C \to B} {\left( v \right)_B}\)each \(v\) in \(V\) , because the vector on the right side of (8) is \((d)\underline {\,\,\,\,\,\,\,\,\,\,\,\,} \).

If \(B = \left\{ {{b_1}.....{b_n}} \right\}\) and \(C = \left\{ {{c_1}.....{c_n}} \right\}\) are the bases for vector space \(V\), then according to the theorem of change of coordinates matrix from \(B\) to \(C\), there is always a matrix \(\mathop P\limits_{C \to B} \) of dimension \(n \times n\) such that \({\left( x \right)_c} = \mathop P\limits_{C \to B} {\left( x \right)_B}\). It means that the vectors in basis \(B\) have \(C\) coordinate vectors, which are the same as the columns of \(\mathop P\limits_{C \to B} \).

As \(B = \left\{ {{b_1}.....{b_n}} \right\}\) and \(C = \left\{ {{c_1}.....{c_n}} \right\}\) are the bases for vector space \(V\), we can write \(v = {x_1}{b_1} + {x_2}{b_2} + .... + {x_n}{b_n}\).Thus, “\(B\) is the basis for vector space \(V\)” fills in blank (a).

As the vectors in basis \(B\) have \(C\) coordinate vectors, which are the same as the columns of\(\mathop P\limits_{C \to B} \), coordinate mapping must be a linear transformation. Thus, “coordinate mapping is a linear transformation” fills in blank (b).

The equation of the form \({\left( v \right)_C} = \left( {{{\left( {{b_1}} \right)}_C}\,\,\,{{\left( {{b_2}} \right)}_C}\,\,\,...\,\,\,\,{{\left( {{b_n}} \right)}_C}} \right)\left( \begin{array}{l}{x_1}\\\,\, \vdots \\{x_n}\end{array} \right)\) is a product of a matrix and a vector. The matrix is \(\left( {{{\left( {{b_1}} \right)}_C}\,\,\,{{\left( {{b_2}} \right)}_C}\,\,\,...\,\,\,\,{{\left( {{b_n}} \right)}_C}} \right)\) , and the vector is \(\left( \begin{array}{l}{x_1}\\\,\, \vdots \\{x_n}\end{array} \right)\).

Thus, “product of matrix and vector” fills in blank (c).

There is always a matrix \(\mathop P\limits_{C \to B} \) of the dimension \(n \times n\) such that \({\left( x \right)_c} = \mathop P\limits_{C \to B} {\left( x \right)_B}\). It implies that the coordinate vector of x relative to \(B\) is changed to \(C\).

So, “the coordinate vector of \(v\) relative to \(B\)” fills in blank (d).