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Q40E

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Linear Algebra and its Applications
Found in: Page 191
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

(M) Let \(H = {\mathop{\rm Span}\nolimits} \left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2}} \right\}\) and \(K = {\mathop{\rm Span}\nolimits} \left\{ {{{\mathop{\rm v}\nolimits} _3},{{\mathop{\rm v}\nolimits} _4}} \right\}\), where

\({{\mathop{\rm v}\nolimits} _1} = \left( {\begin{array}{*{20}{c}}5\\3\\8\end{array}} \right),{{\mathop{\rm v}\nolimits} _2} = \left( {\begin{array}{*{20}{c}}1\\3\\4\end{array}} \right),{{\mathop{\rm v}\nolimits} _3} = \left( {\begin{array}{*{20}{c}}2\\{ - 1}\\5\end{array}} \right),{{\mathop{\rm v}\nolimits} _4} = \left( {\begin{array}{*{20}{c}}0\\{ - 12}\\{ - 28}\end{array}} \right)\)

Then \(H\) and \(K\) are subspaces of \({\mathbb{R}^3}\). In fact, \(H\) and \(K\) are planes in \({\mathbb{R}^3}\) through the origin, and they intersect in a line through 0. Find a nonzero vector w that generates that line. (Hint: w can be written as \({c_1}{{\mathop{\rm v}\nolimits} _1} + {c_2}{{\mathop{\rm v}\nolimits} _2}\) and also as \({c_3}{{\mathop{\rm v}\nolimits} _3} + {c_4}{{\mathop{\rm v}\nolimits} _4}\). To build w, solve the equation \({c_1}{{\mathop{\rm v}\nolimits} _1} + {c_2}{{\mathop{\rm v}\nolimits} _2} = {c_3}{{\mathop{\rm v}\nolimits} _3} + {c_4}{{\mathop{\rm v}\nolimits} _4}\) for the unknown \({c_j}'{\mathop{\rm s}\nolimits} \).)

The nonzero vector w that generates that line is \(\left( {1, - 2, - 1} \right)\).

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Write the augmented matrix

\({\mathop{\rm w}\nolimits} \) can be written as \({c_1}{{\mathop{\rm v}\nolimits} _1} + {c_2}{{\mathop{\rm v}\nolimits} _2}\) and \({c_3}{{\mathop{\rm v}\nolimits} _3} + {c_4}{{\mathop{\rm v}\nolimits} _4}\) because the line lies in both \(H = {\mathop{\rm Span}\nolimits} \left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2}} \right\}\) and \(K = {\mathop{\rm Span}\nolimits} \left\{ {{{\mathop{\rm v}\nolimits} _3},{{\mathop{\rm v}\nolimits} _4}} \right\}\). To determine \({\mathop{\rm w}\nolimits} \), find scalars (c) which solve \({c_1}{{\mathop{\rm v}\nolimits} _1} + {c_2}{{\mathop{\rm v}\nolimits} _2} + {c_3}{{\mathop{\rm v}\nolimits} _3} + {c_4}{{\mathop{\rm v}\nolimits} _4} = 0\).

Consider the augmented matrix shown below:

\(\left( {\begin{array}{*{20}{c}}{{{\mathop{\rm v}\nolimits} _1}}&{{{\mathop{\rm v}\nolimits} _2}}&{ - {{\mathop{\rm v}\nolimits} _3}}&{ - {{\mathop{\rm v}\nolimits} _4}}&0\end{array}} \right) = \left( {\begin{array}{*{20}{c}}5&1&{ - 2}&0&0\\3&3&1&{12}&0\\8&4&{ - 5}&{28}&0\end{array}} \right)\)

Step 2: Convert the matrix into the row-reduced echelon form

Consider the matrix \(A = \left( {\begin{array}{*{20}{c}}5&1&{ - 2}&0&0\\3&3&1&{12}&0\\8&4&{ - 5}&{28}&0\end{array}} \right)\).

Use the code in MATLAB to obtain the row-reduced echelon form as shown below:

\(\begin{array}{l} > > {\mathop{\rm A}\nolimits} = \left( {5\,\,\,1\,\,\, - 2\,\,\,0\,\,\,0;\,3\,\,\,3\,\,\,1\,\,\,\,12\,\,\,0;\,\,\,\,8\,\,\,4\,\,\, - 5\,\,\,28\,\,\,0} \right)\\ > > {\mathop{\rm U}\nolimits} = {\mathop{\rm rref}\nolimits} \left( {\mathop{\rm A}\nolimits} \right)\end{array}\)

\(\left( {\begin{array}{*{20}{c}}5&1&{ - 2}&0&0\\3&3&1&{12}&0\\8&4&{ - 5}&{28}&0\end{array}} \right) \sim \left( {\begin{array}{*{20}{c}}1&0&0&{\frac{{ - 10}}{3}}&0\\0&1&0&{\frac{{26}}{3}}&0\\0&0&1&{ - 4}&0\end{array}} \right)\)

Step 3: Determine the nonzero vector w that generates the line

The vector of \({c_j}'{\mathop{\rm s}\nolimits} \) is a multiple of \(\left( {\frac{{10}}{3},\frac{{ - 26}}{3},4,1} \right)\).

For example, \(\left( {10, - 26,12,3} \right)\) yields \({\mathop{\rm w}\nolimits} = 10{{\mathop{\rm v}\nolimits} _1} - 26{{\mathop{\rm v}\nolimits} _2} = 12{{\mathop{\rm v}\nolimits} _3} + 3{{\mathop{\rm v}\nolimits} _4} = \left( {24, - 48, - 24} \right)\). The other option for \({\mathop{\rm w}\nolimits} \) is \(\left( {1, - 2, - 1} \right)\).

Thus, the nonzero vector w that generates the line is \(\left( {1, - 2, - 1} \right)\).

Most popular questions for Math Textbooks

Given \(T:V \to W\) as in Exercise 35, and given a subspace \(Z\) of \(W\), let \(U\) be the set of all \({\mathop{\rm x}\nolimits} \) in \(V\) such that \(T\left( {\mathop{\rm x}\nolimits} \right)\) is in \(Z\). Show that \(U\) is a subspace of \(V\).

In Exercise 18, A is an \(m \times n\) matrix. Mark each statement True or False. Justify each answer.

18. a. If B is any echelon form of A, then the pivot columns of B form a basis for the column space of A.

b. Row operations preserve the linear dependence relations among the rows of A.

c. The dimension of the null space of A is the number of columns of A that are not pivot columns.

d. The row space of \({A^T}\) is the same as the column space of A.

e. If A and B are row equivalent, then their row spaces are the same.

Question: Determine if the matrix pairs in Exercises 19-22 are controllable.

20. \(A = \left( {\begin{array}{*{20}{c}}{.8}&{ - .3}&0\\{.2}&{.5}&1\\0&0&{ - .5}\end{array}} \right),B = \left( {\begin{array}{*{20}{c}}1\\1\\0\end{array}} \right)\).

Let H be an n-dimensional subspace of an n-dimensional vector space V. Explain why \(H = V\).

Use coordinate vector to test whether the following sets of poynomial span \({{\bf{P}}_{\bf{2}}}\). Justify your conclusions.

a. \({\bf{1}} - {\bf{3}}t + {\bf{5}}{t^{\bf{2}}}\), \( - {\bf{3}} + {\bf{5}}t - {\bf{7}}{t^{\bf{2}}}\), \( - {\bf{4}} + {\bf{5}}t - {\bf{6}}{t^{\bf{2}}}\), \({\bf{1}} - {t^{\bf{2}}}\)

b. \({\bf{5}}t + {t^{\bf{2}}}\), \({\bf{1}} - {\bf{8}}t - {\bf{2}}{t^{\bf{2}}}\), \( - {\bf{3}} + {\bf{4}}t + {\bf{2}}{t^{\bf{2}}}\), \({\bf{2}} - {\bf{3}}t\)
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