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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

Question: Exercises 12-17 develop properties of rank that are sometimes needed in applications. Assume the matrix \(A\) is \(m \times n\).

  1. Show that if \(B\) is \(n \times p\), then rank\(AB \le {\mathop{\rm rank}\nolimits} A\). (Hint: Explain why every vector in the column space of \(AB\) is in the column space of \(A\).
  2. Show that if \(B\) is \(n \times p\), then rank\(AB \le {\mathop{\rm rank}\nolimits} B\). (Hint: Use part (a) to study rank\({\left( {AB} \right)^T}\).)

  1. It is proved that \({\mathop{\rm rank}\nolimits} AB \le {\mathop{\rm rank}\nolimits} A\).
  2. It is proved that \({\mathop{\rm rank}\nolimits} AB \le {\mathop{\rm rank}\nolimits} B\).
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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Show that if \(B\) is \(n \times p\), then rank\(AB \le {\mathop{\rm rank}\nolimits} A\)

a)

According to the theorem 11 let \(H\) be a subspace of a finite-dimensional vector space \(V\), any linearly independent set can be expanded, if necessary, to a basis for \(H\). Also, \(H\) is finite-dimensional such that \(\dim H \le \dim V\).

Consider \({\mathop{\rm y}\nolimits} \) is in \({\mathop{\rm Col}\nolimits} AB\), then \({\mathop{\rm y}\nolimits} = AB\) for some \({\mathop{\rm x}\nolimits} \). However, as \(AB{\mathop{\rm x}\nolimits} = A\left( {B{\mathop{\rm x}\nolimits} } \right)\), \({\mathop{\rm y}\nolimits} = A\left( {B{\mathop{\rm x}\nolimits} } \right)\), and \({\mathop{\rm y}\nolimits} \) is in \({\mathop{\rm Col}\nolimits} A\). Therefore, \({\mathop{\rm Col}\nolimits} AB\) is a subspace of \({\mathop{\rm Col}\nolimits} A\), and \({\mathop{\rm rank}\nolimits} AB = \dim {\mathop{\rm Col}\nolimits} AB \le \dim {\mathop{\rm Col}\nolimits} A = {\mathop{\rm rank}\nolimits} A\) (according to theorem 11).

Thus, it is proved that \({\mathop{\rm rank}\nolimits} AB \le {\mathop{\rm rank}\nolimits} A\).

Step 2: Show that if \(B\) is \(n \times p\), then rank\(AB \le {\mathop{\rm rank}\nolimits} B\)

b)

The rank theorem states that the dimensions of the column space and the row space of a \(m \times n\) matrix \(A\) are equal. This common dimension (the rank of \(A\)) also equals the number of pivot positions in \(A\) and satisfies the equation \({\mathop{\rm rank}\nolimits} A + \dim {\mathop{\rm Nul}\nolimits} A = n\).

\(\begin{array}{c}{\mathop{\rm rank}\nolimits} AB = {\mathop{\rm rank}\nolimits} {\left( {AB} \right)^T}\\ = {\mathop{\rm rank}\nolimits} {B^T}{A^T}\\ \le {\mathop{\rm rank}\nolimits} {B^T}\\ = {\mathop{\rm rank}\nolimits} B\end{array}\)

Thus, it is proved that \({\mathop{\rm rank}\nolimits} AB \le {\mathop{\rm rank}\nolimits} B\).

Most popular questions for Math Textbooks

Let \(A\) be any \(2 \times 3\) matrix such that \({\mathop{\rm rank}\nolimits} A = 1\), let u be the first column of \(A\), and suppose \({\mathop{\rm u}\nolimits} \ne 0\). Explain why there is a vector v in \({\mathbb{R}^3}\) such that \(A = {{\mathop{\rm uv}\nolimits} ^T}\). How could this construction be modified if the first column of \(A\) were zero?

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

22. (M) \(A = \left( {\begin{array}{*{20}{c}}0&1&0&0\\0&0&1&0\\0&0&0&1\\{ - 1}&{ - 13}&{ - 12.2}&{ - 1.5}\end{array}} \right),B = \left( {\begin{array}{*{20}{c}}1\\0\\0\\{ - 1}\end{array}} \right)\).

Question 11: Let \(S\) be a finite minimal spanning set of a vector space \(V\). That is, \(S\) has the property that if a vector is removed from \(S\), then the new set will no longer span \(V\). Prove that \(S\) must be a basis for \(V\).

In Exercise 5, find the coordinate vector \({\left( x \right)_{\rm B}}\) of x relative to the given basis \({\rm B} = \left\{ {{b_{\bf{1}}},...,{b_n}} \right\}\).

5. \({b_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{3}}}\end{array}} \right),{b_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{ - {\bf{5}}}\end{array}} \right),x = \left( {\begin{array}{*{20}{c}}{ - {\bf{2}}}\\{\bf{1}}\end{array}} \right)\)

Question 11: Let \(S\) be a finite minimal spanning set of a vector space \(V\). That is, \(S\) has the property that if a vector is removed from \(S\), then the new set will no longer span \(V\). Prove that \(S\) must be a basis for \(V\).
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