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Chapter 8: Q8.1-15E (page 437)

Question: 15. Let \(A\) be an \({\rm{m}} \times {\rm{n}}\) matrix and, given \({\rm{b}}\) in \({\mathbb{R}^m}\), show that the set \(S\) of all solutions of \(A{\rm{x}} = {\rm{b}}\) is an affine subset of \({\mathbb{R}^n}\).

Short Answer

Expert verified

It is shown that the set \(S\) of all solutions of \(A{\bf{x}} = {\bf{b}}\)is an affine subset of \({\mathbb{R}^n}\).

Step by step solution

01

Describe the given statement

It is given that \(A\) it is a \(m \times n\) matrix, \({\bf{b}} \in {\mathbb{R}^m}\) and \(S\) is the set of all solutions of \(A{\bf{x}} = {\bf{b}}\).

This implies every point in \(S\) satisfies the system \(A{\bf{x}} = {\bf{b}}\), that is, \(S = \left\{ {x:A{\bf{x}} = {\bf{b}}} \right\}\).

02

 Use Theorem 3

According to theorem 3, a nonempty set \(S\) is affine if and only if it is a flat. So, we will have to show that \(S\) is a flat.

Assume that \(W\)is the set of all homogeneous solutions of the system\(A{\rm{x}} = 0\). So, \(W\)must be a subspace of \({\mathbb{R}^n}\).

03

 Use Theorem 6 of solution set of linear systems

If \(p\) be a solution of the system \(A{\bf{x}} = {\bf{b}}\), then the solution set of \(A{\bf{x}} = {\bf{b}}\) is the set of all vectors of the form \(W = p + {v_h}\), where \({v_h}\) is any solution of the homogeneous equation \(A{\bf{x}} = 0\).

Now, as \(S = W + p\), where \(p\) is a solution of the system \(A{\bf{x}} = {\bf{b}}\), so, \(S\) must be a translated set of \(W\). Thus, \(S\) is a flat.

Therefore, the set \(S\) of all solutions of \(A{\bf{x}} = {\bf{b}}\) is an affine subset of \({\mathbb{R}^n}\).

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Most popular questions from this chapter

Question: Let \({\bf{p}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{3}}}\\{\bf{1}}\\{\bf{2}}\end{array}} \right)\), \({\bf{n}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{\bf{1}}\\{\bf{5}}\\{ - {\bf{1}}}\end{array}} \right)\), \({{\bf{v}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{1}}\\{\bf{1}}\\{\bf{1}}\end{array}} \right)\), \({{\bf{v}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{ - {\bf{2}}}\\{\bf{0}}\\{\bf{1}}\\{\bf{3}}\end{array}} \right)\), and \({{\bf{v}}_{\bf{3}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{4}}\\{\bf{0}}\\{\bf{4}}\end{array}} \right)\), and let H be the hyperplane in\({\mathbb{R}^{\bf{4}}}\) with normal n and passing through p. Which of the points \({{\bf{v}}_{\bf{1}}}\), \({{\bf{v}}_{\bf{2}}}\), and \({{\bf{v}}_{\bf{3}}}\) are on the same side of H as the origin, and which are not?

In Exercises 21-26, prove the given statement about subsets A and B of \({\mathbb{R}^n}\), or provide the required example in \({\mathbb{R}^2}\). A proof for an exercise may use results from earlier exercises (as well as theorems already available in the text).

25. \({\mathop{\rm aff}\nolimits} \left( {A \cap B} \right) \subset \left( {{\mathop{\rm aff}\nolimits} A \cap {\mathop{\rm aff}\nolimits} B} \right)\)

Question: 26. Let \({\rm{q}} = \left( \begin{array}{l}2\\3\end{array} \right)\), \({\rm{p}} = \left( \begin{array}{l}6\\1\end{array} \right)\). Find a hyperplane \(\left( {f:d} \right)\) that strictly separates \(B\left( {{\rm{q}},3} \right)\) and \(B\left( {{\rm{p}},1} \right)\).

Question: Let \({{\bf{v}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{0}}\\{\bf{3}}\\{\bf{0}}\end{array}} \right)\), \({{\bf{v}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{ - {\bf{1}}}\\{\bf{0}}\\{\bf{4}}\end{array}} \right)\), and \({{\bf{v}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{ - {\bf{1}}}\\{\bf{2}}\\{\bf{1}}\\{\bf{1}}\end{array}} \right)\)

\({{\bf{p}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{5}}\\{ - {\bf{3}}}\\{\bf{5}}\\{\bf{3}}\end{array}} \right)\) b. \({{\bf{p}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{ - {\bf{9}}}\\{{\bf{10}}}\\{\bf{9}}\\{ - {\bf{13}}}\end{array}} \right)\) c. \({{\bf{p}}_{\bf{3}}} = \left( {\begin{array}{*{20}{c}}{\bf{4}}\\{\bf{2}}\\{\bf{8}}\\{\bf{5}}\end{array}} \right)\)

and \(S = \left\{ {{{\bf{v}}_1},\,\,{{\bf{v}}_2},\,{{\bf{v}}_3}} \right\}\). It can be shown that S is linearly independent.

a. Is \({{\bf{p}}_{\bf{1}}}\) is span S? Is \({{\bf{p}}_{\bf{1}}}\) is \({\bf{aff}}\,S\)?

b. Is \({{\bf{p}}_{\bf{2}}}\) is span S? Is \({{\bf{p}}_{\bf{2}}}\) is \({\bf{aff}}\,S\)?

c. Is \({{\bf{p}}_{\bf{3}}}\) is span S? Is \({{\bf{p}}_{\bf{3}}}\) is \({\bf{aff}}\,S\)?

The 鈥淏鈥� in B-spline refers to the fact that a segment \({\bf{x}}\left( t \right)\)may be written in terms of a basis matrix, \(\,{M_S}\) , in a form similar to a B茅zier curve. That is,

\({\bf{x}}\left( t \right) = G{M_S}{\bf{u}}\left( t \right)\)for \(\,0 \le t \le 1\)

where \(G\) is the geometry matrix \(\,\left( {{{\bf{p}}_{\bf{0}}}\,\,\,\,{{\bf{p}}_{{\bf{1}}\,\,\,}}\,{{\bf{p}}_{\bf{2}}}\,\,\,{{\bf{p}}_{\bf{3}}}} \right)\)and \({\bf{u}}\left( {\bf{t}} \right)\) is the column vector \(\left( {1,\,\,t,\,\,{t^2},\,{t^3}} \right)\) . In a uniform B-spline, each segment uses the same basis matrix \(\,{M_S}\), but the geometry matrix changes. Construct the basis matrix \(\,{M_S}\) for \({\bf{x}}\left( t \right)\).
See all solutions

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