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

Short Answer

Expert verified

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

Step by step solution

01

Write the polynomials in the standard vector form

The vectorsof the given polynomials can be written as follows:

\(1 + 2{t^3} \equiv \left( {\begin{array}{*{20}{c}}1\\0\\0\\2\end{array}} \right)\),\(\left( {2 + t - 3{t^2}} \right) \equiv \left( {\begin{array}{*{20}{c}}2\\1\\{ - 3}\\0\end{array}} \right)\),\(\left( { - t + 2{t^2} - {t^3}} \right) \equiv \left( {\begin{array}{*{20}{c}}0\\{ - 1}\\2\\{ - 1}\end{array}} \right)\)

02

Form the matrix using the vectors

The matrix formed by using the vectors of the polynomials is:

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

03

Write the matrix in the echelon form

\(\left( {\begin{array}{*{20}{c}}1&2&0\\0&1&{ - 1}\\0&{ - 3}&2\\2&0&{ - 1}\end{array}} \right) \sim \left( {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\\0&0&0\end{array}} \right)\)

As the matrix has a pivot in each column, its columns are linearly independent.

So, the polynomials are linearly independent.

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 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

In Exercise 1, find the vector x determined by the given coordinate vector \({\left( x \right)_{\rm B}}\) and the given basis \({\rm B}\).

1. \({\rm B} = \left\{ {\left( {\begin{array}{*{20}{c}}{\bf{3}}\\{ - {\bf{5}}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{ - {\bf{4}}}\\{\bf{6}}\end{array}} \right)} \right\},{\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{\bf{5}}\\{\bf{3}}\end{array}} \right)\)

Exercises 23-26 concern a vector space V, a basis \(B = \left\{ {{{\bf{b}}_{\bf{1}}},....,{{\bf{b}}_n}\,} \right\}\) and the coordinate mapping \({\bf{x}} \mapsto {\left( {\bf{x}} \right)_B}\).

Show that the coordinate mapping is onto \({\mathbb{R}^n}\). That is, given any y in \({\mathbb{R}^n}\), with entries \({y_{\bf{1}}}\),….,\({y_n}\), produce u in V such that \({\left( {\bf{u}} \right)_B} = y\).

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\({\left( {AB} \right)^T}\).)

If the null space of A \({\bf{7}} \times {\bf{6}}\) matrix A is 4-dimensional, what is the dimension of the column space of A?

In Exercise 4, find the vector x determined by the given coordinate vector \({\left( x \right)_{\rm B}}\) and the given basis \({\rm B}\).

4. \({\rm B} = \left\{ {\left( {\begin{array}{*{20}{c}}{ - {\bf{1}}}\\{\bf{2}}\\{\bf{0}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\bf{3}}\\{ - {\bf{5}}}\\{\bf{2}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\bf{4}}\\{ - {\bf{7}}}\\{\bf{3}}\end{array}} \right)} \right\},{\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{ - {\bf{4}}}\\{\bf{8}}\\{ - {\bf{7}}}\end{array}} \right)\)

See all solutions

Recommended explanations on Math Textbooks

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