/*! 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 3 Show that \(\\{(1,2,3),(0,1,2),(... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 3

Show that \(\\{(1,2,3),(0,1,2),(0,0,1)\\}\) is a basis for \(\mathbb{R}^{3}\).

Short Answer

Expert verified
The given vectors \(\{(1,2,3),(0,1,2),(0,0,1)\}\) form a basis for \(\mathbb{R}^{3}\) as they have been shown to be linearly independent and span the vector space \(\mathbb{R}^{3}\).

Step by step solution

01

Check for Linear Independence

To check for linear independence, we need to show that the only solution to the linear combination \(a(1,2,3) + b(0,1,2) + c(0,0,1) = 0\) is for \(a, b, c = 0\). This results in a system of three equations: \(a = 0\), \(2a + b = 0\), and \(3a + 2b + c = 0\). Solving these simultaneously, the only solution is \(a = b = c = 0\), proving that our vectors are linearly independent.
02

Check for Spanning

To show that the given vectors span \(\mathbb{R}^{3}\), we need to confirm that any vector in \(\mathbb{R}^{3}\) can be written as a linear combination of our given vectors. In general, any vector in \(\mathbb{R}^{3}\) can be written as \(v = (x, y, z)\). To form this vector using our given vectors, we need to solve \(a(1,2,3) + b(0,1,2) + c(0,0,1) = (x, y, z)\). This results in a system of three equations which has a solution for all \(x, y, z\), implying the given vectors span \(\mathbb{R}^{3}\).
03

Conclude

As our vectors are both linearly independent and span \(\mathbb{R}^{3}\), they fulfill both requirements to form a basis for \(\mathbb{R}^{3}\). Thus, we've demonstrated that \(\{(1,2,3),(0,1,2),(0,0,1)\}\) is indeed a basis for \(\mathbb{R}^{3}\).

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Independence
The concept of linear independence is fundamental in understanding vector spaces. It serves as a standard to determine if a set of vectors has the unique capability to express other vectors without relying on each other.

Let's break down what this actually means. Imagine you've been given several ropes to hold up a tent. If you can use each rope without needing another to support the same direction, they are linearly independent. Now, translate this idea to vectors: if a set of vectors in \(\mathbb{R}^{3}\) is linearly independent, no vector in the set can be constructed by a combination of the others.

In our exercise, to prove the linear independence of \(\{(1,2,3),(0,1,2),(0,0,1)\}\), we see that none of the vectors can be formed by scaling or adding up the others. This property is crucial for vectors to be part of a basis, as it ensures that each vector brings a unique direction to what spans the space.
Spanning Set
When we refer to a spanning set in \(\mathbb{R}^{3}\), we're talking about a collection of vectors capable of creating any vector in that space through linear combinations. It's as if we have building blocks that can construct any structure within a playground.

In simpler terms, if you're given a box of LEGO, and you can build every possible model out of it, your LEGO pieces span the space of all models.

For our set of vectors \(\{(1,2,3),(0,1,2),(0,0,1)\}\), we showed that they span \(\mathbb{R}^{3}\) since any vector \(v = (x, y, z)\) can be recreated with these vectors. Fundamentally, this ensures that the set of vectors is sufficient and powerful enough to navigate the entire three-dimensional space.
Basis of Vector Space
The concept of a basis combines the ideas of linear independence and spanning into a single, harmonious package. A basis for \(\mathbb{R}^{3}\) is a set of vectors that are both linearly independent and span the entire space. This set serves as the 'coordinate system' for that vector space.

If vectors form a basis for \(\mathbb{R}^{3}\), then any point in this three-dimensional space can be described precisely in terms of these vectors, with no ambiguity or redundancy.

Imagine having a unique address for every location in a city. In our exercise, the given set \(\{(1,2,3),(0,1,2),(0,0,1)\}\) meets the criteria of being linearly independent and spanning \(\mathbb{R}^{3}\), making it a perfect 'address system' for this space. It's the sweet spot that captures the full breadth of the space with the minimum number of vectors, preventing any confusion or overlap.

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

Find a basis for the subspace \(S=\left\\{p \in \mathbb{P}_{3} \mid p(5)=0\right\\}\).

a. Show that \(B=\left\\{\left[\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right] \cdot\left[\begin{array}{ll}0 & 0 \\ 1 & 2\end{array}\right],\left[\begin{array}{ll}0 & 1 \\ 2 & 3\end{array}\right],\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]\right\\}\) is a linearly independent subset of \(M(2,2)\). b. With no further computation, explain why \(B\) is a basis for \(\mathbb{M}(2,2)\). c. Write \(\left[\begin{array}{rr}2 & 3 \\ 0 & -1\end{array}\right]\) as a linear combination of the elements of \(B\). d. Find the coordinate vector of \(\left[\begin{array}{ll}2 & 3 \\ 0 & 1\end{array}\right]\) with respect to \(B\).

Prove that if \(\\{\mathbf{v}, \mathbf{w}\\}\) is a basis for a vector space, then \(\\{\mathbf{v}+\mathbf{w}, \mathbf{w}\\}\) is also a basis for the vector space.

Write the polynomials defined by the following formulas as linear combinations of \(p_{1}, p_{2}, p_{3}\), where $$ p_{1}(x)=x+1, \quad p_{2}(x)=x^{2}+x, \quad p_{3}(x)=x^{3}+x^{2} . $$ a. \(x^{3}+x^{2}+x+1\) b. \(x^{2}+2 x+1\) c. \(4 x^{3}-7 x^{2}-3 x+8\) d. \((x+1)\left(x^{2}-x+1\right)\) e. \((x+1)\left(x^{2}+x\right)\) f. \((x+1)^{3}\)

Suppose \(V\) is an infinite-dimensional vector space. For any integer \(n \geq 0\), show that \(V\) contains a subspace of dimension \(n\).
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Calculus

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

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.