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Q8.1-15E

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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: 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 \({\mathbb{R}^n}\).

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

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: 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, \(S = \left\{ {x:A{\bf{x}} = {\bf{b}}} \right\}\).

Step 2:  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 \({\mathbb{R}^n}\).

Step 3:  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 \({\mathbb{R}^n}\).

Most popular questions for Math Textbooks

Question: 16. Let \({\rm{v}} \in {\mathbb{R}^n}\)and let \(k \in \mathbb{R}\). Prove that \(S = \left\{ {{\rm{x}} \in {\mathbb{R}^n}:{\rm{x}} \cdot {\rm{v}} = k} \right\}\)is an affine subset of \({\mathbb{R}^n}\).

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.

Repeat Exercise 25 with \({v_1} = \left[ {\begin{array}{*{20}{c}}1\\{\bf{2}}\\{ - {\bf{4}}}\end{array}} \right]\), \({v_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}{\bf{8}}\\{\bf{2}}\\{ - {\bf{5}}}\end{array}} \right]\), \({v_{\bf{3}}} = \left[ {\begin{array}{*{20}{c}}{\bf{3}}\\{{\bf{10}}}\\{ - {\bf{2}}}\end{array}} \right]\), \({\bf{a}} = \left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{0}}\\{\bf{8}}\end{array}} \right]\), and \({\bf{b}} = \left[ {\begin{array}{*{20}{c}}{.{\bf{9}}}\\{{\bf{2}}.{\bf{0}}}\\{ - {\bf{3}}.{\bf{7}}}\end{array}} \right]\).

Question: 28. Give an example of a compact set \(A\) and a closed set \(B\) in \({\mathbb{R}^2}\) such that \(\left( {{\rm{conv}}\,A} \right) \cap \left( {{\rm{conv}}\,B} \right) = \emptyset \) but \(A\) and \(B\) cannot be strictly separated by a hyperplane.

In Exercises 7 and 8, find the barycentric coordinates of p with respect to the affinely independent set of points that precedes it.

8. \(\left( {\begin{array}{{}}0\\1\\{ - 2}\\1\end{array}} \right),\left( {\begin{array}{{}}1\\1\\0\\2\end{array}} \right),\left( {\begin{array}{{}}1\\4\\{ - 6}\\5\end{array}} \right)\), \({\mathop{\rm p}\nolimits} = \left( {\begin{array}{{}}{ - 1}\\1\\{ - 4}\\0\end{array}} \right)\)
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