Let \(A = \left\{ {{{\mathop{\rm a}\nolimits} _1},{{\mathop{\rm a}\nolimits} _2},{{\mathop{\rm a}\nolimits} _3}} \right\}\) and \(B = \left\{ {{{\mathop{\rm b}\nolimits} _1},{{\mathop{\rm b}\nolimits} _2},{{\mathop{\rm b}\nolimits} _3}} \right\}\) be bases for a vector space \(V\), and suppose \({{\mathop{\rm a}\nolimits} _1} = 4{b_1} - {b_2}\), \({{\mathop{\rm a}\nolimits} _2} = - {b_1} + {b_2} + {{\mathop{\rm b}\nolimits} _3}\) and \({{\mathop{\rm a}\nolimits} _3} = {b_2} - 2{{\mathop{\rm b}\nolimits} _3}\).

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

b. Find \({\left[ {\mathop{\rm x}\nolimits} \right]_B}\) for \({\mathop{\rm x}\nolimits} = 3{{\mathop{\rm a}\nolimits} _1} + 4{{\mathop{\rm a}\nolimits} _2} + {{\mathop{\rm a}\nolimits} _3}\).

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 \({{\mathop{\rm a}\nolimits} _1} = 4{{\mathop{\rm b}\nolimits} _1} - {{\mathop{\rm b}\nolimits} _2},{{\mathop{\rm a}\nolimits} _2} = - {{\mathop{\rm b}\nolimits} _1} + {{\mathop{\rm b}\nolimits} _2} + {{\mathop{\rm b}\nolimits} _3},\) and \({{\mathop{\rm a}\nolimits} _3} = {{\mathop{\rm b}\nolimits} _2} - 2{{\mathop{\rm b}\nolimits} _3}\). Then, \({\left[ {{{\mathop{\rm a}\nolimits} _1}} \right]_B} = \left[ {\begin{array}{*{20}{c}}4\\{ - 1}\\0\end{array}} \right],{\left[ {{{\mathop{\rm a}\nolimits} _2}} \right]_B} = \left[ {\begin{array}{*{20}{c}}{ - 1}\\1\\1\end{array}} \right],{\left[ {{{\mathop{\rm a}\nolimits} _3}} \right]_B} = \left[ {\begin{array}{*{20}{c}}0\\1\\{ - 2}\end{array}} \right]\).

\(\begin{aligned} \mathop P\limits_{B \leftarrow A} &= \left[ {\begin{array}{*{20}{c}}{{{\left[ {{{\mathop{\rm a}\nolimits} _1}} \right]}_B}}&{{{\left[ {{{\mathop{\rm a}\nolimits} _2}} \right]}_B}}&{{{\left[ {{{\mathop{\rm a}\nolimits} _3}} \right]}_B}}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}4&{ - 1}&0\\{ - 1}&1&1\\0&1&{ - 2}\end{array}} \right]\end{aligned}\)

Thus, the change-of-coordinates matrix from \(A\) to \(B\) is \(\mathop P\limits_{B \leftarrow A} = \left[ {\begin{array}{*{20}{c}}4&{ - 1}&0\\{ - 1}&1&1\\0&1&{ - 2}\end{array}} \right]\).

b)

It is given that \({\mathop{\rm x}\nolimits} = 3{{\mathop{\rm a}\nolimits} _1} + 4{{\mathop{\rm a}\nolimits} _2} + {{\mathop{\rm a}\nolimits} _3}\), then \({\left[ {\mathop{\rm x}\nolimits} \right]_A} = \left[ {\begin{array}{*{20}{c}}3\\4\\1\end{array}} \right]\).

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

\(\begin{aligned} {\left[ {\mathop{\rm x}\nolimits} \right]_B} &= \mathop P\limits_{B \leftarrow A} {\left[ {\mathop{\rm x}\nolimits} \right]_A}\\ &= \left[ {\begin{array}{*{20}{c}}4&{ - 1}&0\\{ - 1}&1&1\\0&1&{ - 2}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}3\\4\\1\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{12 - 4 + 0}\\{ - 3 + 5 + 1}\\{0 + 4 - 2}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}8\\2\\2\end{array}} \right]\end{aligned}\)

Therefore, the \(B - \)coordinate vector is \({\left[ {\mathop{\rm x}\nolimits} \right]_B} = \left[ {\begin{array}{*{20}{c}}8\\2\\2\end{array}} \right]\).