Question

Let \({b_1} = \left[ {\begin{array}{*{20}{c}}{\bf{0}}\\2\end{array}} \right]\), \({b_2} = \left[ {\begin{array}{*{20}{c}}{\bf{2}}\\{\bf{1}}\end{array}} \right]\), \({\bf{x}} = \left[ {\begin{array}{*{20}{c}}{ - {\bf{2}}}\\{\bf{3}}\end{array}} \right]\), \({\bf{y}} = \left[ {\begin{array}{*{20}{c}}{\bf{2}}\\{\bf{4}}\end{array}} \right]\) \({\bf{z}} = \left[ {\begin{array}{*{20}{c}}{ - {\bf{1}}}\\{ - {\bf{2}}.{\bf{5}}}\end{array}} \right]\), and \(B = \left\{ {{b_1},{b_2}} \right\}\). Use the figure to estimate \({\left[ {\bf{x}} \right]_B}\), \({\left[ {\bf{y}} \right]_B}\) and \({\left[ {\bf{z}} \right]_B}\). Confirm your estimates of \({\left[ {\bf{y}} \right]_B}\) and \({\left[ {\bf{z}} \right]_B}\) by using them and \(\left\{ {{b_1},{b_2}} \right\}\) to compute y and z.

Short answer

The vectors are \({\left[ {\bf{x}} \right]_B} = \left[ {\begin{array}{*{20}{c}}2\\{ - 1}\end{array}} \right]\), \[{\left[ {\bf{y}} \right]_B} = \left[ {\begin{array}{*{20}{c}}{1.5}\\1\end{array}} \right]\], and \[{\left[ {\bf{z}} \right]_B} = \left[ {\begin{array}{*{20}{c}}{ - 1}\\{ - .5}\end{array}} \right]\]. The estimations of \[{\left[ {\bf{y}} \right]_B}\], and \[{\left[ {\bf{z}} \right]_B}\] are confirmed.

Step-by-step solution

Definition of coordinate systems

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:

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

Compute the B coordinate vector of w and x

From the figure, the following are the steps required to reach vector x 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\[{\bf{x}} = {\bf{2}}{{\bf{b}}_1}--{{\bf{b}}_2}\].

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

• From the origin, move 1.5 units in the direction of\({{\bf{b}}_1}\).
• Then, move 1 step in the positive direction of\({{\bf{b}}_2}\).

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

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

• From the origin, move 1 unit in the negative direction of\({{\bf{b}}_1}\).
• Then, move .5 steps in the negative direction of\({{\bf{b}}_2}\).

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

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

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

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

\[{\left[ {\bf{y}} \right]_B} = \left[ {\begin{array}{*{20}{c}}{1.5}\\1\end{array}} \right]\]

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

\[{\left[ {\bf{z}} \right]_B} = \left[ {\begin{array}{*{20}{c}}{ - 1}\\{ - .5}\end{array}} \right]\]

Confirm the estimates

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

Use\[{\bf{y}} = {\bf{1}}.{\bf{5}}{{\bf{b}}_1} + {{\bf{b}}_2}\]to confirm the estimate\[{\left[ {\bf{y}} \right]_B}\].

Then, it can be represented as shown below:

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

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

Use\[{\bf{z}} = - {{\bf{b}}_1} - .5{{\bf{b}}_2}\]to confirm the estimate\[{\left[ {\bf{z}} \right]_B}\].

Then, it can be represented as shown below:

\[\begin{array}{l}{\bf{z}} = - 1\left[ {\begin{array}{*{20}{c}}0\\2\end{array}} \right] - .5\left[ {\begin{array}{*{20}{c}}2\\1\end{array}} \right]\\{\bf{z}} = \left[ {\begin{array}{*{20}{c}}0\\{ - 2}\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}1\\{.5}\end{array}} \right]\\{\bf{z}} = \left[ {\begin{array}{*{20}{c}}{ - 1}\\{ - 2.5}\end{array}} \right]\end{array}\]

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