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Chapter 1: Problem 6

Determine whether we obtain a vector space from \(\mathbb{R}^{2}\) with operations defined by $$ \begin{aligned} \left(v_{1}, v_{2}\right)+\left(w_{1}, w_{2}\right) &=\left(v_{1}+w_{1}, v_{2}+w_{2}\right), \\ r\left(v_{1}, v_{2}\right) &=(0,0) . \end{aligned} $$ It is especially interesting to have an example where exactly one of the axioms fails. This indicates that such an axiom is essential in the sense that there is no way to derive it from the other axioms.

Short Answer

Expert verified
The set of vectors does not constitute a vector space since it fails to satisfy the vector space axiom regarding scalar multiplication and the existence of an additive inverse.

Step by step solution

01

Checking the Scalar Multiplication

Scalar multiplication is defined as r(v1,v2)=(0,0). In a vector space, multiplying any vector by scalar 1 should give the same vector. Let's check with the vector (a, b): \n 1*(a, b) should be (a, b). But according to the given definition, 1*(a, b) = (0, 0). Therefore, the property of scalar multiplication is not preserved.
02

Checking the Existence of Additive Identity and Inverse

In a vector space, for a vector v=(v1,v2), there exists an additive identity, usually (0, 0), such that v + (0, 0) = v. According the given definition, the addition operation is preserved. Thus, additive identify exists. \n For additive inverse, for all v=(v1, v2), there should exist -v such that v + -v = (0, 0). However, due to the changed scalar multiplication, there is no way to produce additive inverse, as any real number multiplied by any vector gives (0, 0). Therefore, the existence of additive inverse is not satisfied.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Scalar Multiplication
Scalar multiplication is a fundamental operation in vector spaces that scales a vector by a scalar. In the given exercise, the scalar multiplication is defined peculiarly as \(r(v_1, v_2) = (0, 0)\), which deviates from the standard definition. Usually, when a vector \((v_1, v_2)\) is multiplied by a scalar \(r\), one would expect the result to be \((rv_1, rv_2)\). This operation should satisfy certain properties, such as:\
    \
  • Multiplying a vector by 1 should yield the vector itself, \((1 * (a, b) = (a, b))\).
    • \
  • Multiplying a vector by 0 should give the zero vector \((0,0)\).
    • \
  • Distributive properties such as \(r(u+v)=ru+rv\) should hold.
    • \
\However, in our example, multiplying any vector \((a, b)\) by any scalar r leads to the zero vector \((0, 0)\). This clearly violates the basic property of scalar multiplication, notably that multiplying by 1 should leave the vector unchanged. Such a definition of scalar multiplication causes issues in meeting the criteria set by vector space axioms, particularly affecting the structure of the vector space.
Additive Identity
Additive identity plays a crucial role in the structure of vector spaces. It is the special element that, when added to any vector, does not change that vector. In most vector spaces, this identity is the zero vector \((0, 0)\). Any vector \(v = (v_1, v_2)\) should satisfy the relation \((v_1, v_2) + (0, 0) = (v_1, v_2)\). This must hold true for the vector space to be valid.\
\
In the given example, the addition rule defined is \((v_1, v_2) + (w_1, w_2) = (v_1 + w_1, v_2 + w_2)\). This correctly respects the presence of an additive identity because \((v_1, v_2) + (0, 0) = (v_1, v_2)\) remains unchanged. Thus, despite the unusual scalar multiplication given, the vector space retains its additive identity under normal addition rules.
Additive Inverse
The additive inverse is another essential concept where, for any vector \(v = (v_1, v_2)\), there exists another vector \(-v = (-v_1, -v_2)\) such that adding them results in the zero vector: \(v + (-v) = (0, 0)\). The additive inverse effectively "cancels" the vector.\
\
In typical vector spaces, every vector has an additive inverse. But for this exercise with modified scalar multiplication, finding an actual additive inverse encounters a problem. Due to the rule that scalar multiplication by any real number results in \((0, 0)\), it collapses the structure needed to achieve an inverse through standard means.\
    \
  • Addition appears normal: Addition retains typical characteristics; for any vector \(v\), you expect \(-v\) to exist under normal rules.
    • \
  • Scalar definitions hinder inverses: Since multiplying any vector with any scalar, like -1, results in \((0, 0)\), matching a usual inverse form is impossible.
    • \
\
This breakdown shows the inconsistency introduced by unconventional scalar rules, highlighting how foundational scalar operations are to ensuring other vector space axioms, like inverses, hold firm.

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Most popular questions from this chapter

Let \(f, g\), and \(h\) be functions in \(\mathbb{F}(\mathbb{R})\) defined by \(f(x)=(x+1)^{3}, g(x)=x^{3}+1\), and \(h(x)=x^{2}+x\). Let \(a\) and \(b\) denote scalars. a. Write the identity in the independent variable \(x\) that must hold if the function \(f\) is equal to the function \(a g+b h\). b. Find scalars \(a\) and \(b\) so that \(f=a g+b h\).

Find subspaces of \(\mathrm{R}^{2}\) that are not nested; that is, neither is a subset of the other.

Let \(X\) be any nonempty set and let \(n\) be a positive integer. Let \(V\) denote the set of all functions \(f: X \rightarrow \mathbb{R}^{n}\). Any element of \(V\) corresponds to an ordered list of functions \(f_{i}: X \rightarrow \mathbb{R}\) for \(i=1,2, \ldots, n\) such that \(f(x)=\) \(\left(f_{1}(x), f_{2}(x), \ldots, f_{n}(x)\right)\) for all \(x \in X\). Define addition and scalar multiplication on \(V\) in terms of the corresponding operations on the coordinate functions. Show that with these operations, \(V\) is a vector space.

a. Write down the ways of grouping the sum of five vectors. (Keep the vectors in the same order.) b. How many ways can the sum of six vectors be grouped? c. Can you discover a pattern that would enable you to determine how many ways the sum of \(n\) vectors can be grouped?

For each of the following, either explain why the rule gives a well-defined function or find a number that has two values assigned to it. a. \(f: \mathbb{R} \rightarrow \mathbb{R}\) where \(f(x)\) is the third digit to the right of the decimal point in the decimal expansion of \(x\). b. \(g: \mathbb{R} \rightarrow \mathbb{R}\) defined by $$ g(x)= \begin{cases}0 & \text { if } x \text { is irrational } \\ p+q & \text { if } x=\frac{p}{q}, \text { where } p \text { and } q \text { are integers. }\end{cases} $$ c. \(h: \mathbb{R} \rightarrow \mathbb{R}\) defined by $$ h(x)= \begin{cases}0 & \text { if } x \text { is irrational } \\ m-n & \text { if } x=\frac{2^{m} p}{2^{2} q}, \text { where } p \text { and } q \text { are odd integers and } \\ & m \text { and } n \text { are nonnegative integers. }\end{cases} $$ d. \(s:[-1,1] \rightarrow \mathbb{R}\) where \(y=s(x)\) satisfies the equation \(x^{2}+y^{2}=1\).
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