Question

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 \((d)\underline {\,\,\,\,\,\,\,\,\,\,\,\,} \).

Short answer

(a) \(B\) is the basis for vector space \(V\)

(b) coordinate mapping is a linear transformation

(c) product of matrix and vector

(d) the coordinate vector of \(v\) relative to \(B\)

Step-by-step solution

State the theorem for change of coordinate matrix from \(B\) to \(C\) 

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 \(\mathop P\limits_{C \to B} \).

Use the first line of the statement to fill the gap in (a).

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).

Use the last line of the statement to fill the gap in (b).

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).

Use the definition of the product of matrix and vector

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).

Use the last line of the statement to fill the gap in (d)

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).