/*! 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 7 Show that the element 0 in a vec... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 7

Show that the element 0 in a vector space is unique.

Short Answer

Expert verified
Assuming there are two distinct zero elements in a vector space V, denoted as \(0_1\) and \(0_2\), we showed that adding them together, \(0_1 + 0_2\), using the identity properties of the zero elements, we get both \(0_1\) and \(0_2\) respectively. Comparing these two results, we find that \(0_1 = 0_2\), proving the element 0 in a vector space is unique.

Step by step solution

01

Add the zero elements

Add the two zero elements 0_1 and 0_2 together: \(0_1+ 0_2 = r\)
02

Use the identity property of the first zero element

According to the identity property of a vector space, adding the zero element to any element should result in the same element. Therefore, adding 0_1 to any element should return the same element: \(x + 0_1 = x\) for all \(x \in V\) Now, let's apply this property to our result from Step 1: \(0_2 + 0_1 = r\) Since, \(x + 0_1 = x\), let's replace the \(x\) with \(0_2\): \(0_2 + 0_1 = 0_2\) Thus, we have: \(0_1 + 0_2 = 0_2\)
03

Use the identity property of the second zero element

Similarly, the identity property holds for 0_2 as well: \(x + 0_2 = x\) for all \(x \in V\) Now, let's apply this property to our result from Step 2: \(0_1 + 0_2 = 0_2\) Since, \(x + 0_2 = x\), let's replace the \(x\) with \(0_1\): \(0_1 + 0_2 = 0_1\) Thus, we have: \(0_1 + 0_2 = 0_1\)
04

Compare the results from the previous steps

Now, let's compare the results from Step 2 and Step 3: From Step 2: \(0_1 + 0_2 = 0_2\) From Step 3: \(0_1 + 0_2 = 0_1\) Since both expressions are equal to \(0_1 + 0_2\), we can equate them: \(0_1 = 0_2\)
05

Conclusion

Since we have shown that \(0_1 = 0_2\), it means that there is only one unique element 0 in the vector space V.

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.

Unpacking the Vector Space Concept
A vector space is a fundamental concept in linear algebra, characterized by a collection of elements called vectors, which can be scaled and added together while satisfying specific rules. These rules, known as axioms, ensure that vector operations are consistent and meaningful in the context of the space.

Key properties of vector spaces include closure under addition and scalar multiplication, the existence of an additive identity (the zero vector), and the existence of additive inverses. These properties provide a stable framework for a wide range of mathematical and real-world applications, including physics, engineering, and computer science.

Visualizing a Vector Space

Imagine a vast, infinite plane where you can place a dot representing a vector anywhere on it. No matter where you place another dot or how you scale it, as long as you follow the vector space rules, the results will still belong to this plane. This consistent framework is what makes vector spaces such a powerful concept in linear algebra.
The Identity Property of Vector Addition
The identity property of vector addition is a crucial characteristic of vector spaces, stating that there is a unique vector鈥攌nown as the zero vector鈥攕uch that when it is added to any vector in the space, the result is that original vector. Symbolically, for a vector space V and any vector \(x \in V\), the zero vector \(0 \in V\) satisfies the equation \(x + 0 = x\).

This identity element acts like a neutral ground or a starting point, maintaining the integrity of other vectors during addition operations. It's like having a balance point; adding zero to something never tips the scale鈥攊t keeps things stable and consistent. This property is not just a mathematical curiosity; it forms the backbone for solving equations and manipulating vectors across various branches of science and engineering.
Navigating the Landscape of Linear Algebra Proofs
Proofs in linear algebra serve to verify the truths within the universe of mathematics these truths build on fundamental premises and logical reasoning to establish the validity of statements about vector spaces and other algebraic structures.

Structure of a Linear Algebra Proof

Proofs typically start with given information or assumptions, then proceed through a series of logical steps, each justified by definitions, axioms, theorems, or previously proven results. The objective is to reach a final statement that confirms the initial claim.

In our original exercise, we deduced the uniqueness of the zero vector through a series of logical steps anchored in the identity property of vector addition. This process not only illustrated how to approach a proof but also showcased the rigorous nature of reasoning in linear algebra鈥攁n invaluable skill in mathematics.

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 \(V\) be the set of all ordered pairs of real numbers with addition defined by \\[ \left(x_{1}, x_{2}\right)+\left(y_{1}, y_{2}\right)=\left(x_{1}+y_{1}, x_{2}+y_{2}\right) \\] and scalar multiplication defined by \\[ \alpha \circ\left(x_{1}, x_{2}\right)=\left(\alpha x_{1}, x_{2}\right) \\] Scalar multiplication for this system is defined in an unusual way, and consequently we use the symbol o to avoid confusion with the ordinary scalar multiplication of row vectors. Is \(V\) a vector space with these operations? Justify your answer.

Find the transition matrix representing the change of coordinates on \(P_{3}\) from the ordered basis \(\left[1, x, x^{2}\right]\) to the ordered basis \\[ \left[1,1+x, 1+x+x^{2}\right] \\]

For each of the following matrices, find a basis for the row space, a basis for the column space, and a basis for the null space. (a) \(\left(\begin{array}{lll}1 & 3 & 2 \\ 2 & 1 & 4 \\ 4 & 7 & 8\end{array}\right)\) (b) \(\left(\begin{array}{rrrr}-3 & 1 & 3 & 4 \\ 1 & 2 & -1 & -2 \\ -3 & 8 & 4 & 2\end{array}\right)\) (c) \(\left(\begin{array}{cccc}1 & 3 & -2 & 1 \\ 2 & 1 & 3 & 2 \\ 3 & 4 & 5 & 6\end{array}\right)\)

Prove that any finite set of vectors that contains the zero vector must be linearly dependent.

Let \(S, T,\) and \(U\) be subspaces of a vector space \(V .\) We can form new subspaces using the operations of \(\cap\) and \(+\) defined in Exercises 23 and 25 When we do arithmetic with numbers, we know that the operation of multiplication distributes over the operation of addition in the sense that \\[ a(b+c)=a b+a c \\] It is natural to ask whether similar distributive laws hold for the two operations with subspaces. (a) Does the intersection operation for subspaces distribute over the addition operation? That is does \\[ S \cap(T+U)=(S \cap T)+(S \cap U) \\] (b) Does the addition operation for subspaces distribute over the intersection operation? That is does \\[ S+(T \cap U)=(S+T) \cap(S+U) \\]
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation

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.