Question

\(B = \left\{ {{{\mathop{\rm b}\nolimits} _1},{{\mathop{\rm b}\nolimits} _2Let }} \right\}\)and \(C = \left\{ {{{\mathop{\rm c}\nolimits} _1},{{\mathop{\rm c}\nolimits} _2}} \right\}\) be bases for a vector space \(V\), and suppose \({{\mathop{\rm b}\nolimits} _1} = 6{{\mathop{\rm c}\nolimits} _1} - 2{{\mathop{\rm c}\nolimits} _2}\) and \({{\mathop{\rm b}\nolimits} _2} = 9{{\mathop{\rm c}\nolimits} _1} - 4{{\mathop{\rm c}\nolimits} _2}\).

a. Find the change-of-coordinates matrix from \(B\) to \(C\).

b. Find \({\left[ {\mathop{\rm x}\nolimits} \right]_c}\) for \({\mathop{\rm x}\nolimits} = - 3{{\mathop{\rm b}\nolimits} _1} + 2{{\mathop{\rm b}\nolimits} _2}\). Use part (a).

Short answer

1. The change-of-coordinates matrix is \(\mathop P\limits_{C \leftarrow B} = \left[ {\begin{array}{*{20}{c}}6&9\\{ - 2}&{ - 4}\end{array}} \right]\).
2. The \(C - \)coordinate vector is \({\left[ {\mathop{\rm x}\nolimits} \right]_C} = \left[ {\begin{array}{*{20}{c}}0\\{ - 2}\end{array}} \right]\).

Step-by-step solution

Change-of-coordinate matrix

Theorem 15states that let \(B = \left\{ {{{\mathop{\rm b}\nolimits} _1},...,{{\mathop{\rm b}\nolimits} _n}} \right\}\) and \(C = \left\{ {{{\mathop{\rm c}\nolimits} _1},...,{{\mathop{\rm c}\nolimits} _n}} \right\}\) be the bases of a vector space \(V\). Then, there is a unique \(n \times n\) matrix \(\mathop P\limits_{C \leftarrow B} \) such that \({\left[ {\mathop{\rm x}\nolimits} \right]_C} = \mathop P\limits_{C \leftarrow B} {\left[ {\mathop{\rm x}\nolimits} \right]_B}\).

The columns of \(\mathop P\limits_{C \leftarrow B} \) are the \(C - \)coordinate vectors of the vectors in the basis \(B\). Thus, \(\mathop P\limits_{C \leftarrow B} = \left[ {\begin{array}{*{20}{c}}{{{\left[ {{{\mathop{\rm b}\nolimits} _1}} \right]}_C}}&{{{\left[ {{{\mathop{\rm b}\nolimits} _2}} \right]}_C}}& \cdots &{{{\left[ {{{\mathop{\rm b}\nolimits} _n}} \right]}_C}}\end{array}} \right]\).

Determine the change-of-coordinate matrix from \(B\) to \(C\)

a)

Suppose \({{\mathop{\rm b}\nolimits} _1} = 6{{\mathop{\rm c}\nolimits} _1} - 2{{\mathop{\rm c}\nolimits} _2}\) and \({{\mathop{\rm b}\nolimits} _2} = 9{{\mathop{\rm c}\nolimits} _1} - 4{{\mathop{\rm c}\nolimits} _2}\). Therefore, \({\left[ {{{\mathop{\rm b}\nolimits} _1}} \right]_C} = \left[ {\begin{array}{*{20}{c}}6\\{ - 2}\end{array}} \right],{\left[ {{{\mathop{\rm b}\nolimits} _2}} \right]_C} = \left[ {\begin{array}{*{20}{c}}9\\{ - 4}\end{array}} \right]\).

\(\begin{aligned} \mathop P\limits_{C \leftarrow B} &= \left[ {\begin{array}{*{20}{c}}{{{\left[ {{{\mathop{\rm b}\nolimits} _1}} \right]}_C}}&{{{\left[ {{{\mathop{\rm b}\nolimits} _2}} \right]}_C}}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}6&9\\{ - 2}&{ - 4}\end{array}} \right]\end{aligned}\)

Thus, the change-of-coordinates matrix from \(B\) to \(C\) is \(\mathop P\limits_{C \leftarrow B} = \left[ {\begin{array}{*{20}{c}}6&9\\{ - 2}&{ - 4}\end{array}} \right]\).

Determine \({\left[ {\mathop{\rm x}\nolimits}  \right]_c}\) for \({\mathop{\rm x}\nolimits}  =  - 3{{\mathop{\rm b}\nolimits} _1} + 2{{\mathop{\rm b}\nolimits} _2}\)

b)

Suppose \({\mathop{\rm x}\nolimits} = - 3{{\mathop{\rm b}\nolimits} _1} + 2{{\mathop{\rm b}\nolimits} _2}\), then \({\left[ {\mathop{\rm x}\nolimits} \right]_B} = \left[ {\begin{array}{*{20}{c}}{ - 3}\\2\end{array}} \right]\).

\(\begin{aligned} {\left[ {\mathop{\rm x}\nolimits} \right]_C} &= \mathop P\limits_{C \leftarrow B} {\left[ {\mathop{\rm x}\nolimits} \right]_B}\\ &= \left[ {\begin{array}{*{20}{c}}6&9\\{ - 2}&{ - 4}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{ - 3}\\2\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{ - 18 + 18}\\{6 - 8}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}0\\{ - 2}\end{array}} \right]\end{aligned}\)

Therefore, the \(C - \)coordinate vector is \({\left[ {\mathop{\rm x}\nolimits} \right]_C} = \left[ {\begin{array}{*{20}{c}}0\\{ - 2}\end{array}} \right]\).