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Chapter 4: Q4E (page 191)

In Exercise 4, find the vector x determined by the given coordinate vector \({\left( x \right)_{\rm B}}\) and the given basis \({\rm B}\).

4. \({\rm B} = \left\{ {\left( {\begin{array}{*{20}{c}}{ - {\bf{1}}}\\{\bf{2}}\\{\bf{0}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\bf{3}}\\{ - {\bf{5}}}\\{\bf{2}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\bf{4}}\\{ - {\bf{7}}}\\{\bf{3}}\end{array}} \right)} \right\},{\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{ - {\bf{4}}}\\{\bf{8}}\\{ - {\bf{7}}}\end{array}} \right)\)

Short Answer

Expert verified

Vector \(x = \left( {\begin{array}{*{20}{c}}0\\1\\{ - 5}\end{array}} \right)\)

Step by step solution

01

Use the definition

The coordinates of x relative to basis\({\rm B} = \left\{ {{b_{\bf{1}}},{b_{\bf{2}}},...,{b_n}} \right\}\)are the weights\({c_{\bf{1}}},{c_{\bf{2}}},...,{c_n}\),such that\(x = {c_{\bf{1}}}{b_{\bf{1}}} + {c_{\bf{2}}}{b_{\bf{2}}} + ... + {c_n}{b_n}\). Then,\({\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{{c_1}}\\{{c_2}}\\ \vdots \\{{c_n}}\end{array}} \right)\).

02

Find x

By the above definition, you get

\(\begin{array}{c}x = - 4\left( {\begin{array}{*{20}{c}}{ - 1}\\2\\0\end{array}} \right) + 8\left( {\begin{array}{*{20}{c}}3\\{ - 5}\\2\end{array}} \right) + \left( { - 7} \right)\left( {\begin{array}{*{20}{c}}4\\{ - 7}\\3\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}4\\{ - 8}\\0\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{24}\\{ - 40}\\{16}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{ - 28}\\{49}\\{ - 21}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{4 + 24 - 28}\\{ - 8 - 40 + 49}\\{0 + 16 - 21}\end{array}} \right)\\x = \left( {\begin{array}{*{20}{c}}0\\1\\{ - 5}\end{array}} \right).\end{array}\)

03

Draw a conclusion

Hence, vector \(x = \left( {\begin{array}{*{20}{c}}0\\1\\{ - 5}\end{array}} \right)\).

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Most popular questions from this chapter

Exercises 23-26 concern a vector space V, a basis \(B = \left\{ {{{\bf{b}}_{\bf{1}}},....,{{\bf{b}}_n}\,} \right\}\) and the coordinate mapping \({\bf{x}} \mapsto {\left( {\bf{x}} \right)_B}\).

Show that the coordinate mapping is onto \({\mathbb{R}^n}\). That is, given any y in \({\mathbb{R}^n}\), with entries \({y_{\bf{1}}}\),鈥�.,\({y_n}\), produce u in V such that \({\left( {\bf{u}} \right)_B} = y\).

Use Exercise 28 to explain why the equation\(Ax = b\)has a solution for all\({\rm{b}}\)in\({\mathbb{R}^m}\)if and only if the equation\({A^T}x = 0\)has only the trivial solution.

Exercises 23-26 concern a vector space V, a basis \(B = \left\{ {{{\bf{b}}_{\bf{1}}},....,{{\bf{b}}_n}\,} \right\}\) and the coordinate mapping \({\bf{x}} \mapsto {\left( {\bf{x}} \right)_B}\).

Show that a subset \(\left\{ {{{\bf{u}}_1},...,{{\bf{u}}_p}} \right\}\) in V is linearly independent if and only if the set of coordinate vectors \(\left\{ {{{\left( {{{\bf{u}}_{\bf{1}}}} \right)}_B},.....,{{\left( {{{\bf{u}}_p}} \right)}_B}} \right\}\) is linearly independent in \({\mathbb{R}^n}\)(Hint: Since the coordinate mapping is one-to-one, the following equations have the same solutions, \({c_{\bf{1}}}\),鈥�.,\({c_p}\))

\({c_{\bf{1}}}{{\bf{u}}_{\bf{1}}} + ..... + {c_p}{{\bf{u}}_p} = {\bf{0}}\) The zero vector V

\({\left( {{c_{\bf{1}}}{{\bf{u}}_{\bf{1}}} + ..... + {c_p}{{\bf{u}}_p}} \right)_B} = {\left( {\bf{0}} \right)_B}\) The zero vector in \({\mathbb{R}^n}\)a

(calculus required) Define \(T:C\left( {0,1} \right) \to C\left( {0,1} \right)\) as follows: For f in \(C\left( {0,1} \right)\), let \(T\left( t \right)\) be the antiderivative \({\mathop{\rm F}\nolimits} \) of \({\mathop{\rm f}\nolimits} \) such that \({\mathop{\rm F}\nolimits} \left( 0 \right) = 0\). Show that \(T\) is a linear transformation, and describe the kernel of \(T\). (See the notation in Exercise 20 of Section 4.1.)

Question 11: Let \(S\) be a finite minimal spanning set of a vector space \(V\). That is, \(S\) has the property that if a vector is removed from \(S\), then the new set will no longer span \(V\). Prove that \(S\) must be a basis for \(V\).
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