Let \(D = \left\{ {{{\mathop{\rm d}\nolimits} _1},{{\mathop{\rm d}\nolimits} _2},{{\mathop{\rm d}\nolimits} _3}} \right\}\) and \(F = \left\{ {{{\mathop{\rm f}\nolimits} _1},{{\mathop{\rm f}\nolimits} _2},{{\mathop{\rm f}\nolimits} _3}} \right\}\) be bases for a vector space \(V\), and suppose \({{\mathop{\rm f}\nolimits} _1} = 2{{\mathop{\rm d}\nolimits} _1} - {{\mathop{\rm d}\nolimits} _2} + {{\mathop{\rm d}\nolimits} _3}\), \({{\mathop{\rm f}\nolimits} _2} = 3{{\mathop{\rm d}\nolimits} _2} + {{\mathop{\rm d}\nolimits} _3}\) and \({{\mathop{\rm f}\nolimits} _3} = - 3{{\mathop{\rm d}\nolimits} _1} + 2{{\mathop{\rm d}\nolimits} _3}\).

a. Find the change-of-coordinates matrix from \(F\) to \(D\).

b. Find \({\left[ {\mathop{\rm x}\nolimits} \right]_D}\) for \({\mathop{\rm x}\nolimits} = {{\mathop{\rm f}\nolimits} _1} - 2{{\mathop{\rm f}\nolimits} _2} + 2{{\mathop{\rm f}\nolimits} _3}\).

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 bases of a vector space \(V\). Then according to Theorem 15,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 \(B\). That is, \(\mathop P\limits_{C \leftarrow B} = \left[ {\begin{array}{*{20}{c}}{{{\left[ {{{\mathop{\rm b}\nolimits} _1}} \right]}_C}}&{{{\left[ {{{\mathop{\rm b}\nolimits} _2}} \right]}_C}}& \cdots &{{{\left[ {{{\mathop{\rm b}\nolimits} _n}} \right]}_C}}\end{array}} \right]\).

a)

It is given that \({{\mathop{\rm f}\nolimits} _1} = 2{{\mathop{\rm d}\nolimits} _1} - {{\mathop{\rm d}\nolimits} _2} + {{\mathop{\rm d}\nolimits} _3},{{\mathop{\rm f}\nolimits} _2} = 3{{\mathop{\rm d}\nolimits} _2} + {{\mathop{\rm d}\nolimits} _3}\ and \({{\mathop{\rm f}\nolimits} _3} = - 3{{\mathop{\rm d}\nolimits} _1} + 2{{\mathop{\rm d}\nolimits} _3}\). Then, \({\left[ {{{\mathop{\rm f}\nolimits} _1}} \right]_D} = \left[ {\begin{array}{*{20}{c}}2\\{ - 1}\\1\end{array}} \right],{\left[ {{{\mathop{\rm f}\nolimits} _2}} \right]_D} = \left[ {\begin{array}{*{20}{c}}0\\3\\1\end{array}} \right],{\left[ {{{\mathop{\rm f}\nolimits} _3}} \right]_D} = \left[ {\begin{array}{*{20}{c}}{ - 3}\\0\\2\end{array}} \right]\).

\(\begin{aligned} \mathop P\limits_{D \leftarrow F} &= \left[ {\begin{array}{*{20}{c}}{{{\left[ {{{\mathop{\rm f}\nolimits} _1}} \right]}_D}}&{{{\left[ {{{\mathop{\rm f}\nolimits} _2}} \right]}_D}}&{{{\left[ {{{\mathop{\rm f}\nolimits} _3}} \right]}_D}}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}2&0&{ - 3}\\{ - 1}&3&0\\1&1&2\end{array}} \right]\end{aligned}\)

Thus, the change-of-coordinates matrix from \(F\) to \(D\) is \(\mathop P\limits_{D \leftarrow F} = \left[ {\begin{array}{*{20}{c}}2&0&{ - 3}\\{ - 1}&3&0\\1&1&2\end{array}} \right]\).

b)

It is given that \({\mathop{\rm x}\nolimits} = {{\mathop{\rm f}\nolimits} _1} - 2{{\mathop{\rm f}\nolimits} _2} + 2{{\mathop{\rm f}\nolimits} _3}\), then \({\left[ {\mathop{\rm x}\nolimits} \right]_F} = \left[ {\begin{array}{*{20}{c}}1\\{ - 2}\\2\end{array}} \right]\).

Use part (a) to compute the D-coordinate vector.

\(\begin{aligned} {\left[ {\mathop{\rm x}\nolimits} \right]_D} &= \mathop P\limits_{D \leftarrow F} {\left[ {\mathop{\rm x}\nolimits} \right]_F}\\ &= \left[ {\begin{array}{*{20}{c}}2&0&{ - 3}\\{ - 1}&3&0\\1&1&2\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1\\{ - 2}\\2\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{2 + 0 - 6}\\{ - 1 - 6 + 0}\\{1 - 2 + 4}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{ - 4}\\{ - 7}\\3\end{array}} \right]\end{aligned}\)

Therefore, the \(D - \)coordinate vector is \({\left[ {\mathop{\rm x}\nolimits} \right]_D} = \left[ {\begin{array}{*{20}{c}}{ - 4}\\{ - 7}\\3\end{array}} \right]\).