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Chapter 6: Problem 16

\(\begin{array}{ll}\text { } & \text { By calculating }(1+1)(\mathbf{v}+\mathbf{w}) \text { in two }\end{array}\) ways (using axioms \(\mathrm{S} 2\) and \(\mathrm{S} 3\) ), show that axiom \(\mathrm{A} 2\) follows from the other axioms.

Short Answer

Expert verified
Axiom A2 follows from axioms S2 and S3 by showing both methods yield \(2\mathbf{v} + 2\mathbf{w}\).

Step by step solution

01

Expand the Expression Using S3 First

First, let's use the axiom S3, which states that scalar multiplication is distributive across vector addition: \[a(\mathbf{v} + \mathbf{w}) = a\mathbf{v} + a\mathbf{w}.\]Applying S3 to our expression, we need to calculate \((1+1)(\mathbf{v}+\mathbf{w})\) by distributing the scalar over the addition:\[(1+1)(\mathbf{v} + \mathbf{w}) = (1+1)\mathbf{v} + (1+1)\mathbf{w}.\]
02

Simplify Using Scalar Arithmetic

Now we simplify using the properties of numbers, specifically that \((1+1) = 2\):\[2\mathbf{v} + 2\mathbf\mathbf{w}.\]
03

Expand the Expression Using S2 First

Apply axiom S2, which describes the addition of scalars:\[(a + b)\mathbf{v} = a\mathbf{v} + b\mathbf{v}.\]Apply this method to our given expression:\[(1+1)(\mathbf{v}+\mathbf{w}) = 1(\mathbf{v}+\mathbf{w}) + 1(\mathbf{v}+\mathbf{w}).\]
04

Distribute Each Term Individually Using S3

Now, apply S3 to each term in the expression separately:\[= 1\mathbf{v} + 1\mathbf{w} + 1\mathbf{v} + 1\mathbf{w}.\]
05

Combine Like Terms

Combine like terms:\[= (1\mathbf{v} + 1\mathbf{v}) + (1\mathbf{w} + 1\mathbf{w}).\]This simplifies to:\[= 2\mathbf{v} + 2\mathbf{w}.\]
06

Show Equivalence to Prove A2

Since both methods of calculation result in \(2\mathbf{v} + 2\mathbf{w}\), we have shown that using the axioms S2 and S3 implies the distributivity of the scalar multiplication over the vector sum in a manner that verifies A2. Hence axiom A2, which states that \((a+b)\mathbf{v} = a\mathbf{v} + b\mathbf{v}\), follows from axioms S2 and S3.

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Key Concepts

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

Scalar Multiplication
In vector calculus, scalar multiplication is an operation that involves multiplying a vector by a scalar (a real number). This operation is fundamental because it alters the magnitude of the vector without changing its direction, unless the scalar is negative, which reverses the vector's direction.
  • To multiply a vector by a scalar, you multiply each component of the vector by that scalar.
  • If the vector is \( \mathbf{v} = [v_1, v_2, v_3] \) and the scalar is \( a \), then \( a\mathbf{v} = [a \cdot v_1, a \cdot v_2, a \cdot v_3] \).
This operation scales the vector to a new size, allowing for flexible manipulation in calculations. In the context of the problem, we looked at scalar multiplication as a necessary step to simplify expressions by replicating vector elements. This demonstrates how scalar multiplication can break complex expressions into more manageable parts, like expanding \((1+1)(\mathbf{v} + \mathbf{w})\) through the distributive property to simplify calculations.
Vector Addition
Vector addition is another key concept in vector calculus. It involves combining two or more vectors to produce a resultant vector. This operation is performed by adding the corresponding components of the vectors involved.

When adding two vectors \( \mathbf{v} = [v_1, v_2, v_3] \) and \( \mathbf{w} = [w_1, w_2, w_3] \), the result is a new vector:
  • \( \mathbf{v} + \mathbf{w} = [v_1 + w_1, v_2 + w_2, v_3 + w_3] \).
Vector addition is geometrically equivalent to placing the tail of one vector at the head of the other and is crucial in expressing combined effects in physical systems. In our problem, vector addition appears when expanding expressions like \((1+1)(\mathbf{v} + \mathbf{w})\). Understanding how individual parts are added together lets you confirm facts such as the equality between expressions via operation consistency.
Distributive Property
The distributive property is a fascinating principle that connects scalar multiplication with vector addition. It states that multiplying a scalar with a sum of vectors can be done by distributing the scalar multiplication over each vector individually.

Mathematically, for a scalar \( a \) and vectors \( \mathbf{v} \) and \( \mathbf{w} \), it is expressed as:
  • \( a(\mathbf{v} + \mathbf{w}) = a\mathbf{v} + a\mathbf{w} \).
This property is vital for simplifying vector expressions and was used repeatedly in the solution to simplify and verify expressions. By applying the distributive property to expressions such as \((1+1)(\mathbf{v} + \mathbf{w})\), we can redistribute the scalar across the sum of vectors, confirming consistency in mathematical transformations and proving theorems.

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Most popular questions from this chapter

Are the following sets vector spaces with the indicated operations? If not, why not? a. The set \(V\) of nonnegative real numbers; ordinary addition and scalar multiplication. b. The set \(V\) of all polynomials of degree \(\geq 3\), together with 0 ; operations of \(\mathbf{P}\). c. The set of all polynomials of degree \(\leq 3\); operations of \(\mathbf{P}\). d. The set \(\left\\{1, x, x^{2}, \ldots\right\\} ;\) operations of \(\mathbf{P}\). e. The set \(V\) of all \(2 \times 2\) matrices of the form \(\left[\begin{array}{ll}a & b \\ 0 & c\end{array}\right] ;\) operations of \(\mathbf{M}_{22}\) f. The set \(V\) of \(2 \times 2\) matrices with equal column sums; operations of \(\mathbf{M}_{22}\). g. The set \(V\) of \(2 \times 2\) matrices with zero determinant; usual matrix operations. h. The set \(V\) of real numbers; usual operations. i. The set \(V\) of complex numbers; usual addition and multiplication by a real number. j. The set \(V\) of all ordered pairs \((x, y)\) with the addition of \(\mathbb{R}^{2},\) but using scalar multiplication \(a(x, y)=(a x,-a y)\) \(\mathrm{k}\). The set \(V\) of all ordered pairs \((x, y)\) with the addition of \(\mathbb{R}^{2}\), but using scalar multiplication \(a(x, y)=(x, y)\) for all \(a\) in \(\mathbb{R}\) 1\. The set \(V\) of all functions \(f: \mathbb{R} \rightarrow \mathbb{R}\) with pointwise addition, but scalar multiplication defined by \((a f)(x)=f(a x)\) \(\mathrm{m}\). The set \(V\) of all \(2 \times 2\) matrices whose entries sum to \(0 ;\) operations of \(\mathbf{M}_{22}\). n. The set \(V\) of all \(2 \times 2\) matrices with the addition of \(\mathbf{M}_{22}\) but scalar multiplication \(*\) defined by \(a * X=a X^{T}\).

\(\quad\) Suppose \(V=\operatorname{span}\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}, \ldots, \mathbf{v}_{n}\right\\} .\) If \(\mathbf{u}=a_{1} \mathbf{v}_{1}+a_{2} \mathbf{v}_{2}+\cdots+a_{n} \mathbf{v}_{n}\) where the \(a_{i}\) are in \(\mathbb{R}\) and \(a_{1} \neq 0,\) show that \(V=\operatorname{span}\left\\{\mathbf{u}, \mathbf{v}_{2}, \ldots, \mathbf{v}_{n}\right\\} .\)

a. If \(z\) is not a real number, show that \(\left\\{z, z^{2}\right\\}\) is a basis of the real vector space \(\mathbb{C}\) of all complex numbers. b. If \(z\) is neither real nor pure imaginary, show that \(\\{z, \bar{z}\\}\) is a basis of \(\mathbb{C}\)

If \(z\) is a complex number, show that \(\left\\{z, z^{2}\right\\}\) is independent if and only if \(z\) is not real.

a. Let \(p(x)\) and \(q(x)\) lie in \(\mathbf{P}_{1}\) and suppose that \(p(1) \neq 0, q(2) \neq 0,\) and \(p(2)=0=q(1) .\) Show that \(\\{p(x), q(x)\\}\) is a basis of \(\mathbf{P}_{1}\). [Hint: If \(r p(x)+s q(x)=0,\) evaluate at \(x=1, x=2 .]\) b. Let \(B=\left\\{p_{0}(x), p_{1}(x), \ldots, p_{n}(x)\right\\}\) be a set of polynomials in \(\mathbf{P}_{n}\). Assume that there exist numbers \(a_{0}, a_{1}, \ldots, a_{n}\) such that \(p_{i}\left(a_{i}\right) \neq 0\) for each \(i\) but \(p_{i}\left(a_{j}\right)=0\) if \(i\) is different from \(j .\) Show that \(B\) is a basis of \(\mathbf{P}_{n}\).
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