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Chapter 4: Problem 73

Show that Axiom [ \(\left.\mathrm{A}_{4}\right]\) of a vector space \(V\) (that \(u+v=v+u\) ) can be derived from the other axioms for \(V\).

Short Answer

Expert verified
Given two vectors \(u, v \in V\), we can derive the commutative property of addition (\(u+v=v+u\)) using the other axioms: 1. Use associativity, inverse, and identity element: \((u+v)+(-v) = u+(v+(-v)) \Rightarrow (u+v)+(-v) = u+0 \Rightarrow (u+v)-v = u\). 2. Repeat for \((v+u)\) and \(-u\): \((v+u)+(-u) = v+(u+(-u)) \Rightarrow (v+u)+(-u) = v+0 \Rightarrow (v+u)-u = v\). 3. Equate the expressions: \((u+v)-v=(v+u)-u\), cancel out common elements, and obtain \(u+v=v+u\). Thus, we derived Axiom A4 (\(u+v=v+u\)) using the other axioms of a vector space.

Step by step solution

01

List the other axioms of a vector space

To derive Axiom A4, we will utilize the other axioms of a vector space. They are: 1. Closure under addition: For all \(u, v \in V\), \(u+v\) is an element of \(V\). 2. Associativity of addition: For all \(u, v, w \in V\), \((u+v)+w = u+(v+w)\). 3. Existence of identity element: There exists an element \(0 \in V\) such that for all \(u \in V\), \(u+0 = u\). 4. Existence of inverse elements: For every \(u \in V\), there exists an element \(-u \in V\) such that \(u + (-u) = 0\). 5. Closure under scalar multiplication: For all \(u \in V\) and all scalars \(\alpha\), \(\alpha u\) is an element of \(V\). 6. Distributivity of scalar multiplication: For all \(u, v \in V\) and all scalars \(\alpha,\beta\), \(\alpha(u+v) = \alpha u + \alpha v\) and \((\alpha+\beta)u = \alpha u + \beta u\). 7. Scalar identity: For all \(u \in V\), \(1u = u\), where 1 is the multiplicative identity in the field of scalars. 8. Scalar associativity: For all \(u \in V\) and all scalars \(\alpha, \beta\), \(\alpha(\beta u) = (\alpha \beta) u\). Note that A4 is the commutative property of addition which states that for all \(u, v \in V\), \(u + v = v + u\).
02

Derive A4 using the other axioms

Given any two vectors \(u, v \in V\), we will now derive the commutative property of addition (\(u+v=v+u\)) using the other axioms: 1. Using associativity (Axiom A2): \((u+v)+(-v) = u+(v+(-v))\). 2. Find the inverse: According to Axiom A4, there exists an element \(-v \in V\) such that \(v+(-v)=0\). 3. Apply subtraction operation: \((u+v)+(-v) = u+0\). 4. Apply identity element (Axiom A3): \(u+0 = u\). 5. Substitute back to step 1: \((u+v)+(-v) = u \Rightarrow (u+v)-v = u\). 6. Repeat the above process for \((v+u)\) and \(-u\): - Using associativity (Axiom A2): \((v+u)+(-u) = v+(u+(-u))\). - Apply subtraction operation: \((v+u)+(-u) = v+0\). - Apply identity element (Axiom A3): \(v+0 = v\). - Substitute back to step 6 step 1: \((v+u)+(-u) = v \Rightarrow (v+u)-u = v\). 7. If \((u+v)-v = u\) and \((v+u)-u = v\), then \((u+v)-v=(v+u)-u\). 8. Cancel out the common element on both sides: \(u+v=v+u\). Through the above steps, we successfully derived Axiom A4 (\(u+v=v+u\)) using the other axioms of a vector space, proving that A4 can indeed be derived from these axioms.

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Key Concepts

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

Associativity of Addition
When studying vector spaces, the concept of associativity of addition is fundamental. It refers to the idea that the way in which you group vectors when adding them does not affect the result of the addition. In more technical terms, for any three vectors \(u, v, w \in V\), the axiom states that \( (u+v)+w = u+(v+w) \). This property guarantees that, no matter how we parenthesize our addition, the outcome remains unchanged.
  • This axiom is central when we need to handle lengthy linear combinations of vectors, allowing us to rearrange and simplify expressions without worrying about order.
  • It is also used to prove other properties of vector spaces, like the commutative property of addition, which we can derive by cleverly using associativity along with the existence of inverse elements and the identity element of vector addition.
Understanding the associativity of addition allows for flexibility and confidence when performing operations within a vector space, laying the groundwork for further algebraic manipulation and the proof of other important properties.

Commutative Property
The commutative property is a core axiom in vector spaces which makes calculations intuitive and manageable. It simply states that the order in which two vectors are added does not matter: for all vectors \(u, v \in V\), \(u + v = v + u\). Despite its simplicity, this property is powerful鈥攊t ensures that the structure of the space is ordered in a specific way that adheres to our common understanding of addition.
  • In practical terms, it allows for rearrangement of terms when adding multiple vectors, which is particularly helpful when working with vector equations or simplifying expressions.
  • As shown in the textbook solution, the commutative property can be deduced from other axioms, illustrating how the framework of a vector space is tightly interconnected: axioms don't exist in isolation but are rather part of a unified system.
The solution's step-wise approach nicely illustrates how to leverage concepts like the existence of negative vectors and the additive identity to arrive at the conclusion that vector addition is indeed commutative.

Scalar Multiplication
Scalar multiplication is a cornerstone operation in vector spaces that connects the fields of scalars鈥攖ypically real or complex numbers鈥攚ith the geometric structure of vectors. It is defined as an operation that takes a scalar \( \alpha \) and a vector \(u \in V\) and produces another vector \(\alpha u \in V\), which is a 'scaled' version of \(u\). The set of axioms pertaining to scalar multiplication ensures that:
  • Scalars distribute over vector addition, which means stretching or shrinking the sum of vectors is the same as adding the individually scaled vectors.
  • The multiplicative identity scalar leaves a vector unchanged when used in scalar multiplication, encapsulating the notion that multiplying by one should not alter the object.
  • Scaling operations can be consistently performed in succession due to scalar associativity, meaning the effect of scaling by two scalars in sequence is the same as scaling by their product.
Each axiom associated with scalar multiplication contributes to a robust, predictable framework for studying vectors and their transformations, making scalar multiplication not just an operation but an essential tool for understanding linear algebra and its various applications.

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Most popular questions from this chapter

Prove Theorem 4.16: Let \(V\) be a vector space of finite dimension and let \(S=\left\\{u_{1}, u_{2}, \ldots, u_{r}\right\\}\) be a set of linearly independent vectors in \(V\). Then \(S\) is part of a basis of \(V ;\) that is, \(S\) may be extended to a basis of \(V\).

Show that \(u=(a, b)\) and \(v=(c, d)\) in \(K^{2}\) are linearly dependent if and only if \(a d-b c=0\).

Find a homogeneous system whose solution space is spanned by \(v_{1}=(1,2,1,2,1), \quad v_{2}=(1,3,2,5,3), \quad v_{3}=(1,3,3,6,7)\)

Let \(V\) be the set of ordered pairs \((a, b)\) of real numbers. Show that \(V\) is not a vector space over \(\mathbf{R}\) with addition and scalar multiplication defined by (i) \((a, b)+(c, d)=(a+d, b+c)\) and \(k(a, b)=(k a, k b)\), (ii) \((a, b)+(c, d)=(a+c, b+d)\) and \(k(a, b)=(a, b)\), (iii) \((a, b)+(c, d)=(0,0)\) and \(k(a, b)=(k a, k b)\), (iv) \((a, b)+(c, d)=(a c, b d)\) and \(k(a, b)=(k a, k b)\).

Let \(V\) be the set of infinite sequences \(\left(a_{1}, a_{2}, \ldots\right)\) in a field \(K\). Show that \(V\) is a vector space over \(K\) with addition and scalar multiplication defined by $$ \left(a_{1}, a_{2}, \ldots\right)+\left(b_{1}, b_{2}, \ldots\right)=\left(a_{1}+b_{1}, a_{2}+b_{2}, \ldots\right) \text { and } k\left(a_{1}, a_{2}, \ldots\right)=\left(k a_{1}, k a_{2}, \ldots\right) $$
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