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Q5E

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Linear Algebra and its Applications
Found in: Page 437
Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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Short Answer

Question: In Exercise 5, determine whether or not each set is compact and whether or not it is convex.

5. Use the sets from Exercise 3.

  1. Not compact but convex
  2. Compact and convex
  3. Not compact but convex
  4. Not compact and not convex
  5. Not compact but convex
See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Use Exercise 3(a)

(a)

Form Exercise 3(a), it is obtained that \(\left\{ {\left( {x,y} \right):y > 0} \right\}\) is open.

This set is convex as this is not closed and thus not bounded.

Hence, the set \(\left\{ {\left( {x,y} \right):y > 0} \right\}\) is not compact but convex.

Step 2: Use Exercise 3(b)

(b)

Form Exercise 3(b), it is obtained that the set \(\left\{ {\left( {x,y} \right):x = 2\,\,{\rm{and}}\,\,1 \le y \le 3} \right\}\) is closed.

This set is convex as it is closed and bounded.

Hence, the set \(\left\{ {\left( {x,y} \right):x = 2\,\,{\rm{and}}\,\,1 \le y \le 3} \right\}\) is compact and convex.

Step 3: Use Exercise 3(c)

(c)

Form Exercise 3(c), it is observed that the set \(\left\{ {\left( {x,y} \right):x = 2\,\,{\rm{and}}\,\,1 < y < 3} \right\}\) is neither open nor closed.

This set is convex and bounded.

Hence, \(\left\{ {\left( {x,y} \right):x = 2\,\,{\rm{and}}\,\,1 < y < 3} \right\}\)not compact but convex.

Step 4: Use Exercise 3(d)

(d)

Form Exercise 3(d), it is observed that the set \(\left\{ {\left( {x,y} \right):xy = 1\,\,{\rm{and}}\,\,x > 0} \right\}\) is closed.

This set is not convex and not bounded.

Hence, \(\left\{ {\left( {x,y} \right):y > 0} \right\}\) not compact and not convex.

Step 5: Use Exercise 3(e)

(e)

Form Exercise 3(e), it is observed that the set \(\left\{ {\left( {x,y} \right):xy \ge 1\,\,{\rm{and}}\,\,x > 0} \right\}\) is closed.

This set is convex and not bounded.

Hence, \(\left\{ {\left( {x,y} \right):y > 0} \right\}\) is not compact but convex.

Most popular questions for Math Textbooks

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?

The parametric vector form of a B-spline curve was defined in the Practice Problems as

\({\bf{x}}\left( t \right) = \frac{1}{6}\left[ \begin{array}{l}{\left( {1 - t} \right)^3}{{\bf{p}}_o} + \left( {3t{{\left( {1 - t} \right)}^2} - 3t + 4} \right){{\bf{p}}_1}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \left( {3{t^2}\left( {1 - t} \right) + 3t + 1} \right){{\bf{p}}_2} + {t^3}{{\bf{p}}_3}\end{array} \right]\;\), for \(0 \le t \le 1\) where \({{\bf{p}}_o}\) , \({{\bf{p}}_1}\), \({{\bf{p}}_2}\) , and \({{\bf{p}}_3}\) are the control points.

a. Show that for \(0 \le t \le 1\), \({\bf{x}}\left( t \right)\) is in the convex hull of the control points.

b. Suppose that a B-spline curve \({\bf{x}}\left( t \right)\)is translated to \({\bf{x}}\left( t \right) + {\bf{b}}\) (as in Exercise 1). Show that this new curve is again a B-spline.

A quartic Bézier curve is determined by five control points,

\({{\bf{p}}_{\bf{o}}}{\bf{,}}\,{\rm{ }}{{\bf{p}}_{\bf{1}}}\,{\bf{,}}{\rm{ }}{{\bf{p}}_{\bf{2}}}\,{\bf{,}}{\rm{ }}{{\bf{p}}_{\bf{3}}}\)and \({{\bf{p}}_4}\):

\({\bf{x}}\left( t \right) = {\left( {1 - t} \right)^4}{{\bf{p}}_0} + 4t{\left( {1 - t} \right)^3}{{\bf{p}}_1} + 6{t^2}{\left( {1 - t} \right)^2}{{\bf{p}}_2} + 4{t^3}\left( {1 - t} \right){{\bf{p}}_3} + {t^4}{{\bf{p}}_4}\) for \(0 \le t \le 1\)

Construct the quartic basis matrix \({M_B}\) for \({\bf{x}}\left( t \right)\).

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 \({F_{\bf{1}}} \cap {F_{\bf{2}}}\)?

Question 1: Given points \({{\mathop{\rm p}\nolimits} _1} = \left( {\begin{array}{*{20}{c}}1\\0\end{array}} \right),{\rm{ }}{{\mathop{\rm p}\nolimits} _2} = \left( {\begin{array}{*{20}{c}}2\\3\end{array}} \right),\) and \({{\mathop{\rm p}\nolimits} _3} = \left( {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right)\) in \({\mathbb{R}^2}\), let \(S = {\mathop{\rm conv}\nolimits} \left\{ {{{\mathop{\rm p}\nolimits} _1},{{\mathop{\rm p}\nolimits} _2},{{\mathop{\rm p}\nolimits} _3}} \right\}\). For each linear functional \(f\), find the maximum value \(m\) of \(f\), find the maximum value \(m\) of \(f\) on the set \(S\), and find all points x in \(S\) at which \(f\left( {\mathop{\rm x}\nolimits} \right) = m\).

a.\(f\left( {{x_1},{x_2}} \right) = {x_1} - {x_2}\)

b. \(f\left( {{x_1},{x_2}} \right) = {x_1} + {x_2}\)

c. \(f\left( {{x_1},{x_2}} \right) = - 3{x_1} + {x_2}\)
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