Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Find an example in \({\mathbb{R}^2}\) to show that equality need not hold in the statement of Exercise 23.

Short Answer

Expert verified

It is proved that \({\mathop{\rm aff}\nolimits} \left( {A \cup B} \right) \ne {\mathop{\rm aff}\nolimits} A \cup {\mathop{\rm aff}\nolimits} B\).

Step by step solution

01

Definition of the affine hull

The set of all affine combinations of points in a set\(S\)is called the affine hull(or affine span) of \(S\), denoted by \({\mathop{\rm aff}\nolimits} S\).

02

Find an example in \({\mathbb{R}^2}\) to show that equality need not hold in Exercise 23

RecallTheorem 2,whichstates that a set\(S\)is affineif and only if every affine combination of points of\(S\)lies in\(S\).

That is,\(S\)is affine if and only if\(S = {\mathop{\rm aff}\nolimits} S\).

Recall the equality holds in exercise 23 as shown below:

\(\left( {{\mathop{\rm aff}\nolimits} A} \right) \cup \left( {{\mathop{\rm aff}\nolimits} B} \right) \subset {\mathop{\rm aff}\nolimits} \left( {A \cup B} \right)\)

Let the sets\(A\)and\(B\), each of which contains one or two points.

Consider the set\(A\)containing the point\(\left( {0,0} \right)\), and the set\(B\)containing the point\(\left( {1,1} \right)\).

So,\(B\)contains two points\(\left( {0,0} \right)\)and\(\left( {1,1} \right)\). This implies that\({\mathop{\rm aff}\nolimits} \left( {A \cup B} \right)\)is in the line since it includes two points in lines.

According to theorem 2, if\(A\)are affine, then\(A = {\mathop{\rm aff}\nolimits} A\). Set B is an affine combination of set A since it contains\(\left( {1,1} \right)\).

The\({\mathop{\rm aff}\nolimits} B\)is the set contains two points, where\({\mathop{\rm aff}\nolimits} \left( {A \cup B} \right)\)through\(\left( {0,0} \right)\)but it is not same as\({\mathop{\rm aff}\nolimits} A \cup {\mathop{\rm aff}\nolimits} B\). This implies that\({\mathop{\rm aff}\nolimits} \left( {A \cup B} \right) \ne {\mathop{\rm aff}\nolimits} A \cup {\mathop{\rm aff}\nolimits} B\).

Thus, it is proved that \({\mathop{\rm aff}\nolimits} \left( {A \cup B} \right) \ne {\mathop{\rm aff}\nolimits} A \cup {\mathop{\rm aff}\nolimits} B\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Question: In Exercises 21 and 22, mark each statement True or False. Justify each answer.

22 a. If \(d\)is a real number and \(f\) is a nonzero linear functional defined on \({\mathbb{R}^n}\) , then \(f:d\)is a hyperplane in \({\mathbb{R}^n}\) .

b. Given any vector n and any real number \(d\), the set \(\left\{ {x:n \cdot x = d} \right\}\) is a hyperplane.

c. If \(A\) and \(B\) are nonempty disjoint sets such that \(A\) is compact and \(B\) is closed, then there exists a hyperplane that strictly separates \(A\) and \(B\).

d. If there exists a hyperplane \(H\) such that \(H\) does not strictly separate two sets \(A\) and \(B\), then \(\left( {{\rm{conv}}\,A} \right) \cap \left( {{\rm{conv}}\,B} \right) \ne \emptyset \) and \(B\).

Question: 13. Suppose \(\left\{ {{{\rm{v}}_{\rm{1}}}{\rm{,}}{{\rm{v}}_{\rm{2}}}{\rm{,}}{{\rm{v}}_{\rm{3}}}} \right\}\) is a basis for \({\mathbb{R}^3}\). Show that Span \(\left\{ {{{\rm{v}}_{\rm{2}}} - {{\rm{v}}_{\rm{1}}},{{\rm{v}}_{\rm{3}}} - {{\rm{v}}_{\rm{1}}}} \right\}\) is a plane in \({\mathbb{R}^3}\). (Hint: What can you say about \({\rm{u}}\) and \({\rm{v}}\)when Span \(\left\{ {{\rm{u,v}}} \right\}\) is a plane?)

In Exercises 11 and 12, mark each statement True or False. Justify each answer.

12.a. The essential properties of Bezier curves are preserved under the action of linear transformations, but not translations.

b. When two Bezier curves \({\mathop{\rm x}\nolimits} \left( t \right)\) and \(y\left( t \right)\) are joined at the point where \({\mathop{\rm x}\nolimits} \left( 1 \right) = y\left( 0 \right)\), the combined curve has \({G^0}\) continuity at that point.

c. The Bezier basis matrix is a matrix whose columns are the control points of the curve.

In Exercises 1-4, write y as an affine combination of the other point listed, if possible.

\({{\bf{v}}_{\bf{1}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}\\{\bf{2}}\end{aligned}} \right)\), \({{\bf{v}}_{\bf{2}}} = \left( {\begin{aligned}{*{20}{c}}{ - {\bf{2}}}\\{\bf{2}}\end{aligned}} \right)\), \({{\bf{v}}_{\bf{3}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{0}}\\{\bf{4}}\end{aligned}} \right)\), \({{\bf{v}}_{\bf{4}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{3}}\\{\bf{7}}\end{aligned}} \right)\), \({\bf{y}} = \left( {\begin{aligned}{*{20}{c}}{\bf{5}}\\{\bf{3}}\end{aligned}} \right)\)

Questions: Let \({F_{\bf{1}}}\) and \({F_{\bf{2}}}\) be 4-dimensional flats in \({\mathbb{R}^{\bf{6}}}\), and suppose that \({F_{\bf{1}}} \cap {F_{\bf{2}}} \ne \phi \). What are the possible dimension of \({F_{\bf{1}}} \cap {F_{\bf{2}}}\)?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free