/*! 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 19 Determine whether the given set,... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 19

Determine whether the given set, together with the specified operations of addition and scalar multiplication, is a vector space over the indicated \(\mathbb{Z}_{p} .\) If it is not, list all of the axioms that fail to hold. The set of all vectors in \(\mathbb{Z}_{2}^{n}\) with an odd number of 1s, over \(\mathbb{Z}_{2}\) with the usual vector addition and scalar multiplication.

Short Answer

Expert verified
The set is not a vector space due to failure of closure under addition and scalar multiplication.

Step by step solution

01

Check for Closure under Addition

We need to check whether the sum of any two vectors in the set is also in the set. Vectors in \( \mathbb{Z}_{2}^{n} \) have components 0 or 1, and addition is performed component-wise modulo 2. If both vectors have an odd number of 1s, adding them might result in an even number of 1s due to some 1s cancelling out (1+1=0). If we take an example like \([1, 0]\) and \([1, 1]\), both have an odd number of 1s but their sum \([0, 1]\) has an even number of 1s. Thus it fails closure.
02

Check for Closure under Scalar Multiplication

Vector spaces require that multiplying any vector by a scalar from \( \mathbb{Z}_{2} \), which contains 0 and 1, returns a vector in the set. However, if we multiply any vector by 0, all entries become 0 which has an even number of 1s (since zero ones are present), thereby not in the set. This violates closure under scalar multiplication.
03

List Axioms That Fail

From the previous steps, we can see that the set fails to satisfy: 1) Closure under addition, and 2) Closure under scalar multiplication. These are two essential vector space axioms. The failure of these axioms directly disqualifies the set from being a vector space. We don't need to check the other axioms as these are already not satisfied.

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.

Closure Under Addition
Closure under addition is a fundamental property necessary for a set of vectors to be considered a vector space. This means that if you take any two vectors from the set and add them together, their sum must also be within the same set. In the context of our exercise, we're considering vectors from \( \mathbb{Z}_{2}^{n} \), which have components consisting of only 0s and 1s. The addition of these vectors is performed component-wise, and it follows the rules of arithmetic mod 2.

For a set to have closure under addition, the sum of any two vectors should yield another vector that belongs to the set. However, as seen in examples like \([1, 0]\) and \([1, 1]\), the sum \([0, 1]\) results in an even number of 1s. Since the set is defined to include only vectors with an odd number of 1s, it fails this closure requirement. Hence, the given set of vectors violates closure under addition, making it inconsistent with the requirements of a vector space.
Closure Under Scalar Multiplication
Closure under scalar multiplication is another key criterion for a mathematical structure to be deemed a vector space. It implies that if you scale any vector from the vector set by a scalar, the resulting vector should also reside within the same set. In our exercise, scaling means multiplying each component of the vector by a scalar from \( \mathbb{Z}_{2} \), which is limited to the values 0 and 1.

When a vector is multiplied by the scalar 1, it remains unchanged, retaining the number of 1s in its components. Therefore, if the vector initially had an odd number of 1s, it continues to comply with the set's definition. However, multiplying by 0 results in a zero vector, made entirely of 0s, which inherently contains an even number of 1s (zero being considered even). This multiplication outcome disqualifies the zero vector from the set, demonstrating that the vector set does not maintain closure under scalar multiplication. Consequently, our set fails to satisfy this fundamental vector space property.
Vector Space Axioms
To be a vector space, a set of vectors must satisfy certain axioms or properties, which ensure that vector addition and scalar multiplication behave as expected. These vector space axioms encompass:
  • Associativity of addition
  • Commutativity of addition
  • Identity element of addition
  • Inverse elements of addition
  • Distributive properties for scalars
  • Closure under addition and scalar multiplication
The exercise highlights the violation of two critical axioms: closure under addition and closure under scalar multiplication. Each set failing to meet these conditions cannot be classified as a vector space.

Meeting these axioms ensures that the combined effects of vector addition and scalar multiplication maintain a consistent structure, fitting a vector space's definition. Hence, since the given set fails essential axioms, it disqualifies as a vector space under the given operations over \( \mathbb{Z}_{2} \). Understanding these axioms helps to grasp why not every set can form a vector space and clarifies the strict requirements necessary for such structures.

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 \(a_{0}, a_{1}, \ldots, a_{n}\) be \(n+1\) distinct real numbers. Define \(T: \mathscr{P}_{n} \rightarrow \mathbb{R}^{n+1}\) by \\[ T(p(x))=\left[\begin{array}{c} p\left(a_{0}\right) \\ p\left(a_{1}\right) \\ \vdots \\ p\left(a_{n}\right) \end{array}\right] \\] Prove that \(T\) is an isomorphism.

Find the solution of the differential equation that satisfies the given boundary condition\((s)\) $$f^{\prime \prime}-2 f^{\prime}+5 f=0, f(0)=1, f(\pi / 4)=0$$

Let \(S: V \rightarrow W\) and \(T: U \rightarrow V\) be linear transformations. (a) Prove that if \(S \circ T\) is one-to-one, so is \(T\) (b) Prove that if \(S \circ T\) is onto, so is \(S\).

A pendulum consists of a mass, called a bob, that is affixed to the end of a string of length \(L\) (see Figure 6.26 . When the bob is moved from its rest position and released, it swings back and forth. The time it takes the pendulum to swing from its farthest right position to its farthest left position and back to its next farthest right position is called the period of the pendulum. (Figure can't copy) Let \(\theta=\theta(t)\) be the angle of the pendulum from the vertical. It can be shown that if there is no resistance, then when \(\theta\) is small it satisfies the differential equation \\[ \theta^{\prime \prime}+\frac{g}{L} \theta=0 \\] where \(g\) is the constant of acceleration due to gravity, approximately \(9.7 \mathrm{m} / \mathrm{s}^{2} .\) Suppose that \(L=1 \mathrm{m}\) and that the pendulum is at rest (i.e., \(\theta=0\) ) at time \(t=0\) second. The bob is then drawn to the right at an angle of \(\theta_{1}\) radians and released. (a) Find the period of the pendulum. (b) Does the period depend on the angle \(\theta_{1}\) at which the pendulum is released? This question was posed and answered by Galileo in \(1638 .\) [Galileo Galilei \((1564-1642)\) studied medicine as a student at the University of Pisa, but his real interest was always mathematics. In \(1592,\) Galileo was appointed professor of mathematics at the University of Padua in Venice, where he taught primarily geometry and astronomy. He was the first to use a telescope to look at the stars and planets, and in so doing, he produced experimental data in support of the Copernican view that the planets revolve around the sun and not the earth. For this, Galileo was summoned before the Inquisition, placed under house arrest, and forbidden to publish his results. While under house arrest, he was able to write up his research on falling objects and pendulums. His notes were smuggled out of Italy and published as Discourses on Two New Sciences in \(1638 .\)

Find the solution of the differential equation that satisfies the given boundary condition\((s)\) $$g^{\prime \prime}-2 g=0, g(0)=1, g(1)=0$$
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Logic and Functions

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.