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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.

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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.

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Most popular questions from this chapter

Determine whether the following vectors are linearly independent in \(P_{3}\) (a) \(1, x^{2}, x^{2}-2\) (b) \(2, x^{2}, x, 2 x+3\) (c) \(x+2, x+1, x^{2}-1\) (d) \(x+2, x^{2}-1\)

Let \(S\) be the subspace of \(\mathbb{R}^{2}\) spanned by \(\mathbf{e}_{1}\) and let \(T\) be the subspace of \(\mathbb{R}^{2}\) spanned by \(\mathbf{e}_{2}\). Is \(S \cup T\) a subspace of \(\mathbb{R}^{2} ?\) Explain.

Let \(A\) be a \(5 \times 8\) matrix with rank equal to 5 and let b be any vector in \(\mathbb{R}^{5}\). Explain why the system \(A \mathbf{x}=\mathbf{b}\) must have infinitely many solutions.

Let \(\mathbf{u}_{1}=(1,1,1)^{T}, \mathbf{u}_{2}=(1,2,2)^{T},\) and \(\mathbf{u}_{3}=\) \((2,3,4)^{T}\) (a) Find the transition matrix corresponding to the change of basis from \(\left\\{\mathbf{e}_{1}, \mathbf{e}_{2}, \mathbf{e}_{3}\right\\}\) to \(\left\\{\mathbf{u}_{1}, \mathbf{u}_{2}, \mathbf{u}_{3}\right\\}\) (b) Find the coordinates of each of the following vectors with respect to the ordered basis \(\left\\{\mathbf{u}_{1}, \mathbf{u}_{2}, \mathbf{u}_{3}\right\\}\) (i) \(\quad(3,2,5)^{T}\) (ii) \(\quad(1,1,2)^{T}\) (iii) \((2,3,2)^{T}\)

Show that \(\mathbb{R}^{m \times n},\) together with the usual addition and scalar multiplication of matrices, satisfies the eight axioms of a vector space.
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