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In Exercises 1 and 2, find the vector x determined by the given

coordinate vector \({\left[ {\bf{x}} \right]_{\rm B}}\)and the given basis B. Illustrate your answer with a figure, as in the solution of Practice Problem 2.

2. \(B = \left\{ {\left[ {\begin{array}{*{20}{c}}{ - {\bf{2}}}\\1\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{\bf{3}}\\1\end{array}} \right]} \right\}\), \({\left[ x \right]_B} = \left[ {\begin{array}{*{20}{c}}{ - {\bf{1}}}\\{\bf{3}}\end{array}} \right]\)

Short Answer

Expert verified

The vector is \({\bf{x}} = \left[ {\begin{array}{*{20}{c}}{11}\\2\end{array}} \right]\).

Step by step solution

01

Definition of coordinate systems

For the basis of asubspaceH, 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\(B\)-coordinatevector of x is represented as shown below:

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

02

Construct the augmented matrix

Compare\({\left[ x \right]_B} = \left[ {\begin{array}{*{20}{c}}{ - 1}\\3\end{array}} \right]\)with\({\left[ x \right]_B} = \left[ {\begin{array}{*{20}{c}}{{c_1}}\\{{c_2}}\end{array}} \right]\). So,\({c_1} = - 1\)and\({c_2} = 3\).

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

Consider the vectors\[{{\bf{b}}_1} = \left[ {\begin{array}{*{20}{c}}{ - 2}\\1\end{array}} \right]\],\[{{\bf{b}}_2} = \left[ {\begin{array}{*{20}{c}}3\\1\end{array}} \right]\], and\[{\bf{x}} = \left[ {\begin{array}{*{20}{c}}{ - 1}\\3\end{array}} \right]\].

Then, it can be represented as shown below:

\(\begin{array}{l}{\bf{x}} = - 1\left[ {\begin{array}{*{20}{c}}{ - 2}\\1\end{array}} \right] + 3\left[ {\begin{array}{*{20}{c}}3\\1\end{array}} \right]\\{\bf{x}} = \left[ {\begin{array}{*{20}{c}}2\\{ - 1}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}9\\3\end{array}} \right]\\{\bf{x}} = \left[ {\begin{array}{*{20}{c}}{11}\\2\end{array}} \right]\end{array}\)

Thus, the vector is\({\bf{x}} = \left[ {\begin{array}{*{20}{c}}{11}\\2\end{array}} \right]\).

The illustrated figure is shown below:

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