/*! 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} Problem 26 Extend \(\left\\{u_{1}=(1,1,1,1)... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 26

Extend \(\left\\{u_{1}=(1,1,1,1), u_{2}=(2,2,3,4)\right\\}\) to a basis of \(\mathbf{R}^{4}\).

Short Answer

Expert verified
The basis extended from the given vectors \(\left\\{u_{1}=(1,1,1,1), u_{2}=(2,2,3,4)\right\\}\) is \(\bold{B}=\left\\{u_1=(1,1,1,1), u_2=(2,2,3,4), u_3=(-1,1,0,0), u_4=(-2,-1,0,1)\right\\}\).

Step by step solution

01

Verify that the given vectors are linearly independent

First, we need to make sure that the given vectors \(\left\\{u_{1}=(1,1,1,1), u_{2}=(2,2,3,4)\right\\}\) are linearly independent. To do this, set up the following equation: \[c_1(1,1,1,1)+c_2(2,2,3,4)=(0,0,0,0)\] If the only solution to this equation is \(c_1=0\) and \(c_2=0\), then the given vectors are linearly independent. Now, we can rewrite the equation as a system of linear equations: \[ \begin{cases} c_1+2c_2=0\\ c_1+2c_2=0\\ c_1+3c_2=0\\ c_1+4c_2=0 \end{cases} \] Solving this system, we find that the only solution is \(c_1=0\) and \(c_2=0\), so the given vectors are indeed linearly independent.
02

Find two more linearly independent vectors

To extend the set to a basis, we need to find two additional linearly independent vectors. One way to do this is to set up a matrix with our initial vectors and then complete the matrix to be a 4x4 identity matrix: \[ \begin{pmatrix} 1 & 2 &\star&\star\\ 1 & 2 &\star&\star\\ 1 & 3 &\star&\star\\ 1 & 4 &\star&\star \end{pmatrix} \sim \begin{pmatrix} 1 & 2 & 0 & 0\\ 1 & 2 & 1 & 0\\ 1 & 3 & 0 & 1\\ 1 & 4 & 0 & 0 \end{pmatrix} \] Now, the matrix is in row-echelon form, and we can read off two more linearly independent vectors from the last two columns: \(u_3=(-1,1,0,0)\) and \(u_4=(-2,-1,0,1)\).
03

Combine all the vectors to form a basis

Now that we have found two more linearly independent vectors, we can combine all of them to form a basis for \(\mathbf{R}^4\). Therefore, the completed basis is: \[\bold{B}=\left\\{u_1=(1,1,1,1), u_2=(2,2,3,4), u_3=(-1,1,0,0), u_4=(-2,-1,0,1)\right\\}\] And this completes the problem.

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

Let \(U_{1}, U_{2}, U_{3}\) be the following subspaces of \(\mathbf{R}^{3}\) : \\[U_{1}=\\{(a, b, c): a=c\\}, \quad U_{2}=\\{(a, b, c): a+b+c=0\\}, \quad U_{3}=\\{(0,0, c)\\}\\] Show that \((\mathrm{a}) \mathbf{R}^{3}=U_{1}+U_{2},\) (b) \(\mathbf{R}^{3}=U_{2}+U_{3},(\mathrm{c}) \mathbf{R}^{3}=U_{1}+U_{3} .\) When is the sum direct?

Find a basis and the dimension of the subspace \(W\) of \(\mathbf{P}(t)\) spanned by (a) \(u=t^{3}+2 t^{2}-2 t+1, \quad v=t^{3}+3 t^{2}-3 t+4, \quad w=2 t^{3}+t^{2}-7 t-7\), (b) \(u=t^{3}+t^{2}-3 t+2, \quad v=2 t^{3}+t^{2}+t-4, \quad w=4 t^{3}+3 t^{2}-5 t+2\).

Let the rows of \(A \in M_{n \times n}(F)\) be \(a_{1}, a_{2}, \ldots, a_{n}\), and let \(B\) be the matrix in which the rows are $a_{n}, a_{n-1}, \ldots, a_{1}\(. Calculate \)\operatorname{det}(B)\( in terms of \)\operatorname{det}(A)$.

In Exercises 5- 12, evaluate the determinant of the given matrix by cofactor expansion along the indicated row. $$ \left(\begin{array}{rrr} 0 & 1 & 2 \\ -1 & 0 & -3 \\ 2 & 3 & 0 \end{array}\right) $$ along the first row

Let \(V\) be the set of ordered pairs \((a, b)\) of real numbers with addition in \(V\) and scalar multiplication on \(V\) defined by \\[(a, b)+(c, d)=(a+c, \quad b+d) \quad \text { and } \quad k(a, b)=(k a, 0)\\] Show that \(V\) satisfies all the axioms of a vector space except \(\left[\mathrm{M}_{4}\right]\) -that is, except \(1 u=u\). Hence, \(\left[\mathrm{M}_{4}\right]\) is not a consequence of the other axioms.
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Mechanics Maths

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation

Discrete Mathematics

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.