/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q5E Let \(A = \left\{ {{{\mathop{\rm... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Q5E (page 191)

Let \(A = \left\{ {{{\mathop{\rm a}\nolimits} _1},{{\mathop{\rm a}\nolimits} _2},{{\mathop{\rm a}\nolimits} _3}} \right\}\) and \(B = \left\{ {{{\mathop{\rm b}\nolimits} _1},{{\mathop{\rm b}\nolimits} _2},{{\mathop{\rm b}\nolimits} _3}} \right\}\) be bases for a vector space \(V\), and suppose \({{\mathop{\rm a}\nolimits} _1} = 4{b_1} - {b_2}\), \({{\mathop{\rm a}\nolimits} _2} = - {b_1} + {b_2} + {{\mathop{\rm b}\nolimits} _3}\) and \({{\mathop{\rm a}\nolimits} _3} = {b_2} - 2{{\mathop{\rm b}\nolimits} _3}\).

a. Find the change-of-coordinates matrix from \(A\) to \(B\).

b. Find \({\left[ {\mathop{\rm x}\nolimits} \right]_B}\) for \({\mathop{\rm x}\nolimits} = 3{{\mathop{\rm a}\nolimits} _1} + 4{{\mathop{\rm a}\nolimits} _2} + {{\mathop{\rm a}\nolimits} _3}\).

Short Answer

Expert verified
  1. The change-of-coordinates matrix from \(A\) to \(B\) is \(\mathop P\limits_{B \leftarrow A} = \left[ {\begin{array}{*{20}{c}}4&{ - 1}&0\\{ - 1}&1&1\\0&1&{ - 2}\end{array}} \right]\).
  2. The \(B - \)coordinate vector is \({\left[ {\mathop{\rm x}\nolimits} \right]_B} = \left[ {\begin{array}{*{20}{c}}8\\2\\2\end{array}} \right]\).

Step by step solution

01

State the change-of-coordinate matrix

Let \(B = \left\{ {{{\mathop{\rm b}\nolimits} _1},...,{{\mathop{\rm b}\nolimits} _n}} \right\}\) and \(C = \left\{ {{{\mathop{\rm c}\nolimits} _1},...,{{\mathop{\rm c}\nolimits} _n}} \right\}\) be bases of a vector space \(V\). Then according to Theorem 15,there is a unique \(n \times n\) matrix \(\mathop P\limits_{C \leftarrow B} \) such that \({\left[ {\mathop{\rm x}\nolimits} \right]_C} = \mathop P\limits_{C \leftarrow B} {\left[ {\mathop{\rm x}\nolimits} \right]_B}\).

The columns of \(\mathop P\limits_{C \leftarrow B} \) are the \(C - \)coordinate vectors of the vectors in the basis \(B\). That is, \(\mathop P\limits_{C \leftarrow B} = \left[ {\begin{array}{*{20}{c}}{{{\left[ {{{\mathop{\rm b}\nolimits} _1}} \right]}_C}}&{{{\left[ {{{\mathop{\rm b}\nolimits} _2}} \right]}_C}}& \cdots &{{{\left[ {{{\mathop{\rm b}\nolimits} _n}} \right]}_C}}\end{array}} \right]\).

02

Determine the change-of-coordinate matrix from \(A\) to \(B\)

a)

It is given that \({{\mathop{\rm a}\nolimits} _1} = 4{{\mathop{\rm b}\nolimits} _1} - {{\mathop{\rm b}\nolimits} _2},{{\mathop{\rm a}\nolimits} _2} = - {{\mathop{\rm b}\nolimits} _1} + {{\mathop{\rm b}\nolimits} _2} + {{\mathop{\rm b}\nolimits} _3},\) and \({{\mathop{\rm a}\nolimits} _3} = {{\mathop{\rm b}\nolimits} _2} - 2{{\mathop{\rm b}\nolimits} _3}\). Then, \({\left[ {{{\mathop{\rm a}\nolimits} _1}} \right]_B} = \left[ {\begin{array}{*{20}{c}}4\\{ - 1}\\0\end{array}} \right],{\left[ {{{\mathop{\rm a}\nolimits} _2}} \right]_B} = \left[ {\begin{array}{*{20}{c}}{ - 1}\\1\\1\end{array}} \right],{\left[ {{{\mathop{\rm a}\nolimits} _3}} \right]_B} = \left[ {\begin{array}{*{20}{c}}0\\1\\{ - 2}\end{array}} \right]\).

\(\begin{aligned} \mathop P\limits_{B \leftarrow A} &= \left[ {\begin{array}{*{20}{c}}{{{\left[ {{{\mathop{\rm a}\nolimits} _1}} \right]}_B}}&{{{\left[ {{{\mathop{\rm a}\nolimits} _2}} \right]}_B}}&{{{\left[ {{{\mathop{\rm a}\nolimits} _3}} \right]}_B}}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}4&{ - 1}&0\\{ - 1}&1&1\\0&1&{ - 2}\end{array}} \right]\end{aligned}\)

Thus, the change-of-coordinates matrix from \(A\) to \(B\) is \(\mathop P\limits_{B \leftarrow A} = \left[ {\begin{array}{*{20}{c}}4&{ - 1}&0\\{ - 1}&1&1\\0&1&{ - 2}\end{array}} \right]\).

03

Determine \({\left[ {\mathop{\rm x}\nolimits}   \right]_B}\) for \({\mathop{\rm x}\nolimits}  = 3{{\mathop{\rm a}\nolimits} _1} + 4{{\mathop{\rm a}\nolimits} _2} + {{\mathop{\rm a}\nolimits} _3}\)

b)

It is given that \({\mathop{\rm x}\nolimits} = 3{{\mathop{\rm a}\nolimits} _1} + 4{{\mathop{\rm a}\nolimits} _2} + {{\mathop{\rm a}\nolimits} _3}\), then \({\left[ {\mathop{\rm x}\nolimits} \right]_A} = \left[ {\begin{array}{*{20}{c}}3\\4\\1\end{array}} \right]\).

Use part (a) to compute the B-coordinate vector.

\(\begin{aligned} {\left[ {\mathop{\rm x}\nolimits} \right]_B} &= \mathop P\limits_{B \leftarrow A} {\left[ {\mathop{\rm x}\nolimits} \right]_A}\\ &= \left[ {\begin{array}{*{20}{c}}4&{ - 1}&0\\{ - 1}&1&1\\0&1&{ - 2}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}3\\4\\1\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{12 - 4 + 0}\\{ - 3 + 5 + 1}\\{0 + 4 - 2}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}8\\2\\2\end{array}} \right]\end{aligned}\)

Therefore, the \(B - \)coordinate vector is \({\left[ {\mathop{\rm x}\nolimits} \right]_B} = \left[ {\begin{array}{*{20}{c}}8\\2\\2\end{array}} \right]\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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

15. Let\(A\)be an\(m \times n\)matrix, and let\(B\)be a\(n \times p\)matrix such that\(AB = 0\). Show that\({\mathop{\rm rank}\nolimits} A + {\mathop{\rm rank}\nolimits} B \le n\). (Hint: One of the four subspaces\({\mathop{\rm Nul}\nolimits} A\),\({\mathop{\rm Col}\nolimits} A,\,{\mathop{\rm Nul}\nolimits} B\), and\({\mathop{\rm Col}\nolimits} B\)is contained in one of the other three subspaces.)

In Exercise 6, find the coordinate vector of x relative to the given basis \({\rm B} = \left\{ {{b_{\bf{1}}},...,{b_n}} \right\}\).

6. \({b_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{2}}}\end{array}} \right),{b_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{5}}\\{ - {\bf{6}}}\end{array}} \right),x = \left( {\begin{array}{*{20}{c}}{\bf{4}}\\{\bf{0}}\end{array}} \right)\)

In statistical theory, a common requirement is that a matrix be of full rank. That is, the rank should be as large as possible. Explain why an m n matrix with more rows than columns has full rank if and only if its columns are linearly independent.

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

14. Show that if \(Q\) is an invertible, then \({\mathop{\rm rank}\nolimits} AQ = {\mathop{\rm rank}\nolimits} A\). (Hint: Use Exercise 13 to study \({\mathop{\rm rank}\nolimits} {\left( {AQ} \right)^T}\).)

In Exercise 7, find the coordinate vector \({\left( x \right)_{\rm B}}\) of x relative to the given basis \({\rm B} = \left\{ {{b_{\bf{1}}},...,{b_n}} \right\}\).

7. \({b_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{1}}}\\{ - {\bf{3}}}\end{array}} \right),{b_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{ - {\bf{3}}}\\{\bf{4}}\\{\bf{9}}\end{array}} \right),{b_{\bf{3}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{ - {\bf{2}}}\\{\bf{4}}\end{array}} \right),x = \left( {\begin{array}{*{20}{c}}{\bf{8}}\\{ - {\bf{9}}}\\{\bf{6}}\end{array}} \right)\)
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.