Qis any matrix such that

\({\left( {\bf{v}} \right)_C} = Q{\left( {\bf{v}} \right)_B}\)for each v in V (9)

Set \({\bf{v}} = {{\bf{b}}_{\bf{1}}}\) in (9). Then (9) shows that \({\left( {{{\bf{b}}_{\bf{1}}}} \right)_C}\) is the first column of Q because (a) _____. Similarly, for \(k = {\bf{2}}\),…..n the k^th column of Q is (b) _____ because (c) _____. This shows the matrix \(\mathop P\limits_{C \leftarrow B} \) defined by (5) in Theorem 15 is the only matrix that satisfies condition (4).

The columns of Q are C coordinate vectors of the vectors in basis B,that is,

\(Q = \left( {{{\left( {{{\bf{b}}_1}} \right)}_C}{{\left( {{{\bf{b}}_2}} \right)}_C}.....{{\left( {{{\bf{b}}_k}} \right)}_C}} \right)\).

If you set \({\bf{v}} = {{\bf{b}}_1}\) in (1), then \({\left( {{{\bf{b}}_1}} \right)_C}\) represents the first column of Q because

\(\begin{aligned} {\left( {{{\bf{b}}_1}} \right)_C} &= Q{\left( {{{\bf{b}}_1}} \right)_B}\\ &= \left( {{{\left( {{{\bf{b}}_1}} \right)}_C}\,{{\left( {{{\bf{b}}_2}} \right)}_C}\,\,....\,\,{{\left( {{{\bf{b}}_k}} \right)}_C}} \right)\left( {\begin{array}{*{20}{c}}1\\0\\ \vdots \\0\end{array}} \right)\\ &= Q{{\bf{e}}_1}.\end{aligned}\)

For \(k = 2\),…n, the k^thcolumn of Q is

\(\begin{aligned} {\left( {{{\bf{b}}_k}} \right)_C} &= Q{\left( {{{\bf{b}}_1}} \right)_B}\\ &= \left( {{{\left( {{{\bf{b}}_1}} \right)}_C}\,\,{{\left( {{{\bf{b}}_2}} \right)}_C}\,\,...\,\,{{\left( {{{\bf{b}}_k}} \right)}_C}} \right)\left( {\begin{array}{*{20}{c}}0\\0\\ \vdots \\1\end{array}} \right)\\ &= Q{e_k}.\end{aligned}\)

So, the k^th column of Q is \({\left( {{{\bf{b}}_k}} \right)_C}\).

\(\mathop P\limits_{C + B} = \left( {{{\left( {{{\bf{b}}_1}} \right)}_C}\,\,{{\left( {{{\bf{b}}_2}} \right)}_C}\,\,....\,\,{{\left( {{{\bf{b}}_n}} \right)}_C}} \right)\)is the only matrix that satisfies the condition

\({\left( {\bf{x}} \right)_c} = \mathop P\limits_{C = S} {\left( {\bf{x}} \right)_S}\).

It means,

\(\begin{aligned}{c}{\left( {{{\bf{b}}_k}} \right)_C} &= Q{\left( {{{\bf{b}}_1}} \right)_B}\\ &= \left( {{{\left( {{{\bf{b}}_1}} \right)}_C}\,{{\left( {{{\bf{b}}_2}} \right)}_C}\,...\,\,{{\left( {{{\bf{b}}_k}} \right)}_C}} \right)\left( {\begin{array}{*{20}{c}}0\\0\\ \vdots \\1\end{array}} \right)\\ &= Q{{\bf{e}}_k}.\end{aligned}\)

So, \({\left( {{{\bf{b}}_k}} \right)_C} = Q{{\bf{e}}_k}\).