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Chapter 8: Q5E (page 437)

In Exercises 5 and 6, let \({{\bf{b}}_{\bf{1}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}\\{\bf{1}}\\{\bf{1}}\end{aligned}} \right)\), \({{\bf{b}}_{\bf{2}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}\\{\bf{0}}\\{ - {\bf{2}}}\end{aligned}} \right)\), and \({{\bf{b}}_{\bf{3}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}\\{ - {\bf{5}}}\\{\bf{1}}\end{aligned}} \right)\) and \(S = \left\{ {{{\bf{b}}_{\bf{1}}},\,{{\bf{b}}_{\bf{2}}},\,{{\bf{b}}_{\bf{3}}}} \right\}\). Note that S is an orthogonal basis of \({\mathbb{R}^{\bf{3}}}\). Write each of the given points as an affine combination of the points in the set S, if possible. (Hint: Use Theorem 5 in section 6.2 instead of row reduction to find the weights.)

a. \({{\bf{p}}_{\bf{1}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{3}}\\{\bf{8}}\\{\bf{4}}\end{aligned}} \right)\)

b. \({{\bf{p}}_{\bf{2}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{6}}\\{ - {\bf{3}}}\\{\bf{3}}\end{aligned}} \right)\)

c. \({{\bf{p}}_{\bf{3}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{0}}\\{ - {\bf{1}}}\\{ - {\bf{5}}}\end{aligned}} \right)\)

Short Answer

Expert verified

a. \({{\bf{p}}_1} = 3\left( {{{\bf{b}}_1}} \right) - 1\left( {{{\bf{b}}_2}} \right) - 1\left( {{{\bf{b}}_3}} \right) \in {\rm{aff}}\,\,S\), the sum of coefficients of in S 1.

b. \({{\bf{p}}_2} = 2\left( {{{\bf{b}}_1}} \right) + 0\left( {{{\bf{b}}_2}} \right) + 1\left( {{{\bf{b}}_3}} \right) \notin {\rm{aff}}\,S\), the sum of coefficients of in S is not 1.

c. \({{\bf{p}}_3} = - 1\left( {{{\bf{b}}_1}} \right) + 2\left( {{{\bf{b}}_2}} \right) + 0\left( {{{\bf{b}}_3}} \right) \in {\rm{aff}}\,S\), the sum of coefficients of in S 1.

Step by step solution

01

Find the augmented matrix

Write the augmented matrix by using the given points as shown below:

\(\begin{aligned}{c}M = \left( {\begin{aligned}{*{20}{c}}{{{\bf{b}}_1}}&{{{\bf{b}}_2}}&{{{\bf{b}}_3}}&{{{\bf{p}}_1}}&{{{\bf{p}}_2}}&{{{\bf{p}}_3}}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}2&1&2&3&6&0\\1&0&{ - 5}&8&{ - 3}&{ - 1}\\1&{ - 2}&1&4&3&{ - 5}\end{aligned}} \right)\end{aligned}\)

02

Obtain the row reduced form of the augmented matrix

Write the augmented matrix as shown below:

\(\begin{aligned}{c}M = \left( {\begin{aligned}{*{20}{c}}0&1&{12}&{ - 13}&{12}&2\\1&0&{ - 5}&8&{ - 3}&{ - 1}\\0&{ - 2}&6&{ - 4}&6&{ - 4}\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ \begin{aligned}{l}{R_1} \to {R_1} - 2{R_2}\\{R_3} \to {R_3} - {R_2}\end{aligned} \right\}\\ = \left( {\begin{aligned}{*{20}{c}}1&0&{ - 5}&8&{ - 3}&{ - 1}\\0&1&{12}&{ - 13}&{12}&2\\0&{ - 2}&6&4&6&{ - 4}\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {{R_1} \leftrightarrow {R_2}} \right\}\\ = \left( {\begin{aligned}{*{20}{c}}1&0&{ - 5}&8&{ - 3}&{ - 1}\\0&1&{12}&{ - 13}&{12}&2\\0&1&{ - 3}&2&{ - 3}&2\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {{R_3} \to - \frac{1}{2}{R_3}} \right\}\\ = \left( {\begin{aligned}{*{20}{c}}1&0&{ - 5}&8&{ - 3}&{ - 1}\\0&1&{12}&{ - 13}&{12}&2\\0&0&{ - 15}&{15}&{ - 15}&0\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {{R_3} \to {R_3} - {R_2}} \right\}\\ = \left( {\begin{aligned}{*{20}{c}}1&0&{ - 5}&8&{ - 3}&{ - 1}\\0&1&{12}&{ - 13}&{12}&2\\0&0&1&{ - 1}&1&0\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {{R_3} \to - \frac{1}{{15}}{R_3}} \right\}\\ = \left( {\begin{aligned}{*{20}{c}}1&0&0&3&2&{ - 1}\\0&1&0&{ - 1}&0&2\\0&0&1&{ - 1}&1&0\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ \begin{aligned}{l}{R_1} \to {R_1} + 5{R_3}\\{R_2} \to {R_2} - 12{R_3}\end{aligned} \right\}\end{aligned}\)

03

Check for the affine combination of \({{\bf{p}}_{\bf{1}}}\)

Use the augmented matrix \({{\bf{p}}_1}\) that can be expressed as shown below:

\({{\bf{p}}_1} = 3\left( {{{\bf{b}}_1}} \right) - 1\left( {{{\bf{b}}_2}} \right) - 1\left( {{{\bf{b}}_3}} \right)\)

The sum of coefficients is \(3 - 1 - 1 = 1\).

Thus, \({{\bf{p}}_1}\) is an affine combination of point in set S.

\({{\bf{p}}_1} = 3\left( {{{\bf{b}}_1}} \right) - 1\left( {{{\bf{b}}_2}} \right) - 1\left( {{{\bf{b}}_3}} \right)\)

04

Check for the affine combination of \({{\bf{p}}_{\bf{2}}}\)

Use the augmented matrix \({{\bf{p}}_2}\) that can be expressed as shown below:

\({{\bf{p}}_2} = 2\left( {{{\bf{b}}_1}} \right) + 0\left( {{{\bf{b}}_2}} \right) + 1\left( {{{\bf{b}}_3}} \right)\)

The sum of the coefficients is \(2 + 0 + 1 = 3 \ne 1\).

So, \({{\bf{p}}_2}\) is not an affine combination of point in set S.

05

Check for the affine combination of \({{\bf{p}}_{\bf{3}}}\)

Use the augmented matrix \({{\bf{p}}_3}\) that can be expressed as shown below:

\({{\bf{p}}_3} = - 1\left( {{{\bf{b}}_1}} \right) + 2\left( {{{\bf{b}}_2}} \right) + 0\left( {{{\bf{b}}_3}} \right)\)

The sum of the coefficients is \( - 1 + 2 + 0 = 1\).

So, \({{\bf{p}}_3}\) is an affine combination of point in set S.

\({{\bf{p}}_3} = - 1\left( {{{\bf{b}}_1}} \right) + 2\left( {{{\bf{b}}_2}} \right) + 0\left( {{{\bf{b}}_3}} \right)\)

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Most popular questions from this chapter

Let \({\bf{x}}\left( t \right)\) and \({\bf{y}}\left( t \right)\) be cubic Bézier curves with control points \(\left\{ {{{\bf{p}}_{\bf{o}}}{\bf{,}}{{\bf{p}}_{\bf{1}}}{\bf{,}}{{\bf{p}}_{\bf{2}}}{\bf{,}}{{\bf{p}}_{\bf{3}}}} \right\}\)and \(\left\{ {{{\bf{p}}_{\bf{3}}}{\bf{,}}{{\bf{p}}_{\bf{4}}}{\bf{,}}{{\bf{p}}_{\bf{5}}}{\bf{,}}{{\bf{p}}_{\bf{6}}}} \right\}\) respectively, so that \({\bf{x}}\left( t \right)\) and \({\bf{y}}\left( t \right)\) are joined at \({{\bf{p}}_3}\) . The following questions refer to the curve consisting of \({\bf{x}}\left( t \right)\) followed by \(y\left( t \right)\). For simplicity, assume that the curve is in \({\mathbb{R}^2}\).

a. What condition on the control points will guarantee that the curve has \({C^1}\) continuity at \({{\bf{p}}_3}\) ? Justify your answer.

b. What happens when \({\bf{x'}}\left( 1 \right)\) and \({\bf{y'}}\left( 1 \right)\) are both the zero vector?

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

21. a. A linear transformation from\(\mathbb{R}\)to\({\mathbb{R}^n}\)is called a linear functional.

b. If\(f\)is a linear functional defined on\({\mathbb{R}^n}\), then there exists a real number\(k\)such that\(f\left( x \right) = kx\)for all\(x\)in\({\mathbb{R}^n}\).

c. If a hyper plane strictly separates sets\(A\)and\(B\), then\(A \cap B = \emptyset \)

d. If\(A\)and\(B\)are closed convex sets and\(A \cap B = \emptyset \), then there exists a hyper plane that strictly separate\(A\)and\(B\).

Show that a set\(\left\{ {{{\bf{v}}_{\bf{1}}},...,{{\bf{v}}_p}} \right\}\)in\({\mathbb{R}^{\bf{n}}}\)is affinely dependent when \(p \ge n + 2\).

Question: Suppose that the solutions of an equation \(A{\bf{x}} = {\bf{b}}\) are all of the form \({\bf{x}} = {x_{\bf{3}}}{\bf{u}} + {\bf{p}}\), where \({\bf{u}} = \left( {\begin{array}{*{20}{c}}{\bf{5}}\\{\bf{1}}\\{ - {\bf{2}}}\end{array}} \right)\) and \({\bf{p}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{3}}}\\{\bf{4}}\end{array}} \right)\). Find points \({{\bf{v}}_{\bf{1}}}\) and \({{\bf{v}}_{\bf{2}}}\) such that the solution set of \(A{\bf{x}} = {\bf{b}}\) is \({\bf{aff}}\left\{ {{{\bf{v}}_{\bf{1}}},\,{{\bf{v}}_{\bf{2}}}} \right\}\).

Let\(\left\{ {{p_1},{p_2},{p_3}} \right\}\)be an affinely dependent set of points in\({\mathbb{R}^{\bf{n}}}\)and let\(f:{\mathbb{R}^{\bf{n}}} \to {\mathbb{R}^{\bf{m}}}\)be a linear transformation. Show that\(\left\{ {f\left( {{{\bf{p}}_1}} \right),f\left( {{{\bf{p}}_2}} \right),f\left( {{{\bf{p}}_3}} \right)} \right\}\)is affinely dependent in\({\mathbb{R}^{\bf{m}}}\).
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