Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Q35E (page 191)

Let V be a vector space that contains a linearly independent set \(\left\{ {{u_{\bf{1}}},{u_{\bf{2}}},{u_{\bf{3}}},{u_{\bf{4}}}} \right\}\). Describe how to construct a set of vectors \(\left\{ {{v_{\bf{1}}},{v_{\bf{2}}},{v_{\bf{3}}},{v_{\bf{4}}}} \right\}\) in V such that \(\left\{ {{v_{\bf{1}}},{v_{\bf{3}}}} \right\}\) is a basis for Span\(\left\{ {{v_{\bf{1}}},{v_{\bf{2}}},{v_{\bf{3}}},{v_{\bf{4}}}} \right\}\).

Short Answer

Expert verified

If \({v_1}\) and \({v_3}\) are linearly independent and \({v_2} = a{v_1} + b{v_3}\) and \({v_4} = c{v_1} + d{v_3}\), wherea, b, c, and d are scalar, then \(\left\{ {{v_1},{v_3}} \right\}\) is a basis for \({\rm{Span}}\left\{ {{v_1},{v_2},{v_3},{v_4}} \right\}\).

Step by step solution

01

Construct the linearly independent vectors

Let \({v_1}\) and \({v_3}\) be linearly independent.

02

Construct a relation between the vectors, except \({v_{\bf{1}}}\) and

Let \({v_2} = a{v_1} + b{v_3}\) and \({v_4} = c{v_1} + d{v_3}\), wherea, b, c, and d are scalar.

By the spanning set theorem, \({\rm{Span}}\left\{ {{v_1},{v_2},{v_3},{v_4}} \right\} = {\rm{Span}}\left\{ {{v_1},{v_3}} \right\}\).

03

Draw a conclusion

In this way, \(\left\{ {{v_1},{v_3}} \right\}\) forms a basis for \({\rm{Span}}\left\{ {{v_1},{v_2},{v_3},{v_4}} \right\}\).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

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?

In Exercises 27-30, use coordinate vectors to test the linear independence of the sets of polynomials. Explain your work.

\({\bf{1}} - {\bf{2}}{t^{\bf{2}}} - {t^{\bf{3}}}\), \(t + {\bf{2}}{t^{\bf{3}}}\), \({\bf{1}} + t - {\bf{2}}{t^{\bf{2}}}\)

If the null space of a\({\bf{5}} \times {\bf{6}}\)matrix A is 4-dimensional, what is the dimension of the row space of A?

Is it possible that all solutions of a homogeneous system of ten linear equations in twelve variables are multiples of one fixed nonzero solution? Discuss.

In Exercises 27-30, use coordinate vectors to test the linear independence of the sets of polynomials. Explain your work.
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Statistics

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Theoretical and Mathematical Physics

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.

Sign-up for free