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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 additive identities, 0 and 0', in a vector space, we can add 0 to the equation involving 0': \(v + 0' = v\). Using the associative law of addition and substituting the additive identity property, we find that \(v + 0' = v\). This shows that 0 and 0' have the exact same effect on any element v in the vector space, making them indistinguishable from one another. Therefore, only one unique additive identity element exists in a vector space, proving that the element 0 in a vector space is unique.

Step by step solution

01

Definition of a vector space

A vector space is a set V together with operations of vector addition and scalar multiplication that satisfy a number of axioms (properties). These properties include closure under addition and scalar multiplication, commutative and associative laws for addition, existence of additive identity, and existence of additive inverses.
02

Additive identity property

One of the properties of vector spaces is the existence of an additive identity鈥攁n element 0 in the vector space such that for any other element, v in V, the following is true: \(v + 0 = v\)
03

Assume two additive identities exist

Let's assume that there are two additive identity elements in the vector space, 0 and 0'. That means both of these elements satisfy the additive identity property: 1. \(v + 0 = v\) 2. \(v + 0' = v\)
04

Prove the uniqueness of the additive identity

To prove that 0 and 0' are actually the same element and that the additive identity is unique in the vector space, let's add 0 to the equation involving 0': \(v + 0' = v\) Add 0 to both sides: \(0 + (v + 0') = 0 + v\) By the associative law of addition, we can rearrange the parentheses: \((0 + v) + 0' = v + 0\) Now substitute the additive identity property (1) to get: \(v + 0' = v\) Since we assumed that 0' was also an additive identity, this equation is equivalent to our second assumption. This means that 0' and 0 have the exact same effect on any element v in the vector space. Thus, since 0 and 0' are indistinguishable from one another, only one unique additive identity element exists in the vector space. The element 0 in a vector space is unique.

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Most popular questions from this chapter

Let \(\mathbb{R}^{+}\) denote the set of positive real numbers. Define the operation of scalar multiplication, denoted ?, by $$\alpha \circ x=x^{\alpha}$$ for each \(x \in \mathbb{R}^{+}\) and for any real number \(\alpha\). Define the operation of addition, denoted \(\oplus,\) by $$x \oplus y=x \cdot y \quad \text { for all } \quad x, y \in \mathbb{R}^{+}$$ Thus, for this system, the scalar product of -3 \(\operatorname{times} \frac{1}{2}\) is given by $$-3 \circ \frac{1}{2}=\left(\frac{1}{2}\right)^{-3}=8$$ and the sum of 2 and 5 is given by $$2 \oplus 5=2 \cdot 5=10$$ Is \(\mathbb{R}^{+}\) a vector space with these operations? Prove your answer.

Let \(\mathbf{x}_{1}, \mathbf{x}_{2},\) and \(\mathbf{x}_{3}\) be linearly independent vectors in \(\mathbb{R}^{n}\) and let $$\mathbf{y}_{1}=\mathbf{x}_{1}+\mathbf{x}_{2}, \quad \mathbf{y}_{2}=\mathbf{x}_{2}+\mathbf{x}_{3}, \quad \mathbf{y}_{3}=\mathbf{x}_{3}+\mathbf{x}_{1}$$ Are \(\mathbf{y}_{1}, \mathbf{y}_{2},\) and \(\mathbf{y}_{3}\) linearly independent? Prove your answer.

Let \(S\) denote the set of all infinite sequences of real numbers with scalar multiplication and addition defined by $$\begin{aligned} \alpha\left\\{a_{n}\right\\} &=\left\\{\alpha a_{n}\right\\} \\ \left\\{a_{n}\right\\}+\left\\{b_{n}\right\\} &=\left\\{a_{n}+b_{n}\right\\} \end{aligned}$$ Show that \(S\) is a vector space.

Show that if a matrix \(U\) is in row echelon form, then the nonzero row vectors of \(U\) form a basis for the row space of \(U\).

Consider the vectors \\[ \mathbf{x}_{1}=\left(\begin{array}{l} 2 \\ 1 \end{array}\right), \quad \mathbf{x}_{2}=\left[\begin{array}{l} 4 \\ 3 \end{array}\right], \quad \mathbf{x}_{3}=\left[\begin{array}{r} 7 \\ -3 \end{array}\right] \\] (a) Show that \(x_{1}\) and \(x_{2}\) form a basis for \(\mathbb{R}^{2}\) (b) Why must \(\mathbf{x}_{1}, \mathbf{x}_{2},\) and \(\mathbf{x}_{3}\) be linearly dependent? (c) What is the dimension of \(\operatorname{Span}\left(\mathbf{x}_{1}, \mathbf{x}_{2}, \mathbf{x}_{3}\right) ?\)
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