/*! 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 18 Prove that every vector space ha... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 18

Prove that every vector space has a basis.

Short Answer

Expert verified
To prove that every vector space has a basis, consider any non-empty vector space V, and start with a linearly independent subset, say {v_1}. If the vector space is not spanned, add a new vector v_i from V that is not already in the span of the current linearly independent set. Continue adding new vectors to the set and maintain linear independence until the vector space is spanned. This process shows that every vector space, whether finite-dimensional or infinite-dimensional, has a basis.

Step by step solution

01

Start with a non-empty vector space

Let V be any non-empty vector space.
02

Take a linearly independent subset

Pick a non-zero vector v_1 from the set V. Since {v_1} is a linearly independent set, we can use it as a starting point.
03

Add new vectors and check for linear independence

Add a new vector v_i from V such that v_i is not in the span of the current set of linearly independent vectors {v_1, v_2, ..., v_{i-1}}. We can now check if this new set of vectors {v_1, v_2, ..., v_i} remains linearly independent: Consider the equation: \(\alpha_1v_1 + \alpha_2v_2 + ... + \alpha_{i-1}v_{i-1} + \alpha_iv_i = 0\) We want to show that if this equation holds, then all the scalars (\(\alpha_1,\alpha_2,...,\alpha_i\)) must be equal to 0. Since \(v_i\) is not in the span of {v_1, v_2, ..., v_{i-1}}, it cannot be written as a linear combination of these vectors. Therefore, \(\alpha_i\) must be 0. Similarly, by the linear independence of the set {v_1, v_2, ..., v_{i-1}}, we can deduce that the other scalars (\(\alpha_1,\alpha_2,...,\alpha_{i-1}\)) must also be equal to 0. This means that the new set {v_1, v_2, ..., v_i} remains linearly independent.
04

Repeat the process

Continue adding new vectors from V to the set of linearly independent vectors until the vector space is spanned. If V is finite-dimensional, this process will eventually end with a finite basis. If V is infinite-dimensional, this process will continue infinitely, but the vector space will still be spanned by the infinite basis. In either case, we have shown that every vector space has a basis.

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.

Basis
In the world of linear algebra, a *basis* is an important concept that simplifies working with vector spaces. Think of a basis as a special set of vectors that provide a convenient way to describe every vector in the space. These vectors are often referred to as *basis vectors*. A basis has two crucial properties:

  • They must be linearly independent.
  • They must span the vector space.
To put it simply, you can uniquely express any vector in the space by a combination of these basis vectors. If you need to describe vectors to someone else, just giving them a list of basis vectors and how much of each one you need is enough. This makes bases a fundamental tool in mathematics, physics, and engineering. Having a basis ensures that you can "build" the entire space from these vectors alone.
Linear Independence
The concept of *linear independence* is essential when talking about sets of vectors. A set of vectors is said to be linearly independent if no vector in the set can be represented as a linear combination of the others. In other words, you can't "make" one vector using the others.

To check for linear independence of a set like \( \{v_1, v_2, ..., v_i\} \), you consider the equation:\[\alpha_1v_1 + \alpha_2v_2 + \ldots + \alpha_iv_i = 0\]where \( \alpha_1, \alpha_2, ..., \alpha_i \) are scalars. The set is linearly independent if the only solution to this equation is given by all \( \alpha \)'s being zero.

Linear independence guarantees that each vector in the set contributes something "new" to the span of the set, meaning no vector is redundant. It's a key property ensuring a well-defined and efficient basis for a vector space.
Finite-Dimensional Vector Space
A *finite-dimensional vector space* is a vector space that can be spanned by a finite number of vectors. This implies that there exists a finite basis for the vector space. In practical terms, these spaces are easier to work with because they can be fully described with a limited number of vectors.

Common examples include the plane, which can be described with just two vectors, or our 3-dimensional space, which needs three basis vectors. Once a basis is established, any vector in the space can be expressed as a combination of the basis vectors. This property makes calculations and various operations simpler and easily manageable.

Finite-dimensional spaces are heavily utilized in computer graphics, engineering, and real-world physics because they align well with the dimensionality of our physical world.
Infinite-Dimensional Vector Space
An *infinite-dimensional vector space* is a bit more abstract compared to its finite-dimensional counterpart. These spaces require an infinite number of vectors to form a basis, meaning no finite set of vectors will be enough to cover the entire space.

You can think of infinite-dimensional spaces as having infinitely more detail or complexity, much like how a line has infinite points. Common examples include the vector spaces of functions, such as those involving polynomials or sequences. Here, each "point" might represent a whole function, rather than a simple number.

Dealing with infinite-dimensional spaces is crucial in fields such as quantum mechanics and certain aspects of functional analysis, where these concepts help explain phenomena that can't be captured by just three or even four dimensions alone.

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

(a) Show that the Hilbert space \(L^{2}([0,1])\) is separable. (b) Show that the Hilbert space \(L^{2}(\mathbf{R})\) is separable. (c) Show that the Banach space \(\ell^{\infty}\) is not separable.

Prove that if \(\mu\) is a measure and \(f, g \in L^{2}(\mu),\) then $$ \|f\|^{2}\|g\|^{2}-|\langle f, g\rangle|^{2}=\frac{1}{2} \iint|f(x) g(y)-g(x) f(y)|^{2} d \mu(y) d \mu(x). $$

Suppose \(f\) and \(g\) are elements of an inner product space. Prove that \(\langle f, g\rangle=0\) if and only if $$ \|f\| \leq\|f+\alpha g\| $$ for all \(\alpha \in \mathbf{F}\).

Find \(a, b \in \mathbf{R}^{3}\) such that \(a\) is a scalar multiple of \((1,6,3), b\) is orthogonal to \((1,6,3),\) and \((5,4,-2)=a+b\).

Suppose \(U\) is a closed subspace of a Hilbert space \(V\) and \(f \in V\). Prove that \(\left\|P_{U} f\right\| \leq\|f\|,\) with equality if and only if \(f \in U\) [This exercise asks you to prove \(8.37(d) .]\)
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Applied Mathematics

Read Explanation

Geometry

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Calculus

Read Explanation

Probability and Statistics

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.