Let \({b_1} = \left[ {\begin{array}{*{20}{c}}3\\0\end{array}} \right]\), \({b_2} = \left[ {\begin{array}{*{20}{c}}{ - {\bf{1}}}\\{\bf{2}}\end{array}} \right]\), \({\bf{w}} = \left[ {\begin{array}{*{20}{c}}{\bf{7}}\\{ - {\bf{2}}}\end{array}} \right]\), \({\bf{x}} = \left[ {\begin{array}{*{20}{c}}{\bf{4}}\\{\bf{1}}\end{array}} \right]\), and \(B = \left\{ {{b_1},{b_2}} \right\}\). Use the figure to estimate \({\left[ {\bf{w}} \right]_B}\) and \({\left[ {\bf{x}} \right]_B}\). Confirm your estimate of \({\left[ {\bf{x}} \right]_B}\) by using it and \(\left\{ {{b_1},{b_2}} \right\}\)to compute x.

For the basis of asubspace H, let the set be\(B = \left\{ {{{\bf{b}}_1},...,{{\bf{b}}_p}} \right\}\). For theweights \({c_1},...,{c_p}\), the coordinate of x is represented as\({\bf{x}} = {c_1}{{\bf{b}}_1} + \cdots + {c_p}{{\bf{b}}_p}\).

The\(B\)-coordinate vector of x is represented as shown below:

\({\left[ {\bf{x}} \right]_B} = \left[ {\begin{array}{*{20}{c}}{{c_1}}\\ \vdots \\{{c_p}}\end{array}} \right]\)

From the figure, the following are the steps required to reach vector w, as shown below:

• From the origin, move 2 units in the direction of\({{\bf{b}}_1}\).
• Then, move 1 step in the negative direction of\({{\bf{b}}_2}\).

In the vector form, it is represented as\[{\bf{w}} = {\bf{2}}{{\bf{b}}_1}--{{\bf{b}}_2}\].

The following are the steps required to reach vector x as shown below:

• From the origin, move 1.5 units in the direction of\({{\bf{b}}_1}\).
• Then, move .5 step in the positive direction of\({{\bf{b}}_2}\).

In the vector form, it is represented as\[{\bf{x}} = {\bf{1}}.{\bf{5}}{{\bf{b}}_1} + .{\bf{5}}{{\bf{b}}_2}\].

The\(B\)-coordinate vectorof x is represented as shown below:

\({\left[ {\bf{x}} \right]_B} = \left[ {\begin{array}{*{20}{c}}{1.5}\\{.5}\end{array}} \right]\)

The\(B\)-coordinate vector of w is represented as shown below:

\[{\left[ {\bf{w}} \right]_B} = \left[ {\begin{array}{*{20}{c}}2\\{ - 1}\end{array}} \right]\]

Compare\({\left[ {\bf{x}} \right]_B} = \left[ {\begin{array}{*{20}{c}}{{c_1}}\\{{c_2}}\end{array}} \right]\)with \({\left[ {\bf{x}} \right]_B} = \left[ {\begin{array}{*{20}{c}}{1.5}\\{.5}\end{array}} \right]\). So,\({c_1} = 1.5\), and\({c_2} = .5\).

For theweights\({c_1}\)and\({c_2}\), and the vector x in H, it must satisfy the equation\({\bf{x}} = {c_1}{{\bf{b}}_1} + {c_2}{{\bf{b}}_2}\).

Consider the vectors\({b_1} = \left[ {\begin{array}{*{20}{c}}3\\0\end{array}} \right]\)and\[{{\bf{b}}_2} = \left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right]\].

Then, it can be represented as shown below:

\(\begin{array}{l}{\bf{x}} = 1.5\left[ {\begin{array}{*{20}{c}}3\\0\end{array}} \right] + .5\left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right]\\{\bf{x}} = \left[ {\begin{array}{*{20}{c}}{4.5}\\0\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - .5}\\1\end{array}} \right]\\{\bf{x}} = \left[ {\begin{array}{*{20}{c}}4\\1\end{array}} \right]\end{array}\)

Thus, the estimation of \({\left[ {\bf{x}} \right]_B}\) is confirmed.