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Chapter 20: Problem 10

Let \(V\) be a vector space over \(F\). Prove that \(-(\alpha v)=(-\alpha) v=\alpha(-v)\) for all \(\alpha \in F\) and all \(v \in V\).

Short Answer

Expert verified
Question: Prove the identity \(-(\alpha v)=(-\alpha) v=\alpha(-v)\) for all scalars \(\alpha\) and all vectors \(v\).

Step by step solution

01

Understand the notation and definitions

Here, \(V\) represents a vector space over the field \(F\), and \(\alpha \in F\) stands for a scalar value, while \(v \in V\) represents a vector. In vector spaces, there are certain properties that the elements must satisfy such as the existence of additive inverses for every element. In this context, the additive inverse of a scalar multiple of a vector (\(\alpha v\)) must also satisfy these properties.
02

Prove \(-(\alpha v)\) is the additive inverse of \(\alpha v\)

We will first show that \(-(\alpha v)\) is the additive inverse of \(\alpha v\). By definition, an additive inverse must satisfy the following property: \(v + (-v) = 0\), where \(0\) is the zero vector in the vector space \(V\). Now, we have \((\alpha v) + (-(\alpha v))\). By the distributive property of scalar multiplication over vector addition, we can write this as: \(\alpha(v + (-v)) = \alpha0=0\) Since \(v + (-v) = 0\), the equation simplifies to \(\alpha0=0\). Therefore, we have shown that \(-(\alpha v)\) is the additive inverse of \(\alpha v\).
03

Prove \((-\alpha)v\) is the additive inverse of \(\alpha v\)

Next, we will show that \((-\alpha)v\) is the additive inverse of \(\alpha v\). We proceed similarly to the previous step, considering the sum \((\alpha v) + ((-\alpha)v)\). By the distributive property of scalar multiplication over vector addition, we can write this as: \((\alpha + (-\alpha))v = 0v\) Since \(\alpha + (-\alpha) = 0\), the equation simplifies to \(0v=0\). Therefore, we have shown that \((-\alpha)v\) is also the additive inverse of \(\alpha v\).
04

Prove \(\alpha(-v)\) is the additive inverse of \(\alpha v\)

Finally, we will show that \(\alpha(-v)\) is the additive inverse of \(\alpha v\). We proceed similarly to the previous steps, considering the sum \((\alpha v) + (\alpha(-v))\). By the distributive property of scalar multiplication over vector addition, we can write this as: \(\alpha(v + (-v)) = \alpha0=0\) Since \(v + (-v) = 0\), the equation simplifies to \(\alpha0=0\). Therefore, we have shown that \(\alpha(-v)\) is also the additive inverse of \(\alpha v\).
05

Establish the identity

Since all three expressions \(-(\alpha v)\), \((-\alpha)v\), and \(\alpha(-v)\) are additive inverses of \(\alpha v\), they must be equal to each other according to the uniqueness of additive inverses in a vector space. Thus, we have established the identity: \(-(\alpha v)=(-\alpha)v=\alpha(-v)\) for all \(\alpha \in F\) and all \(v \in V\).

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Key Concepts

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

Vector Addition
Vector addition is a fundamental operation in vector spaces. It involves taking two vectors and combining them to produce a third vector. You can think of this like adding points or directions together.
For example, if you have two vectors, such as \( \mathbf{a} \) and \( \mathbf{b} \), their vector addition \( \mathbf{a} + \mathbf{b} \) will result in a new vector.
  • This operation is commutative, meaning \( \mathbf{a} + \mathbf{b} = \mathbf{b} + \mathbf{a} \).
  • It's also associative, so \( (\mathbf{a} + \mathbf{b}) + \mathbf{c} = \mathbf{a} + (\mathbf{b} + \mathbf{c}) \).
These properties ensure that vector addition is predictable and uniform, no matter which vectors you're adding. Vector addition is crucial in defining many other vector space concepts because it forms the basis of how vectors interact within the space.
Scalar Multiplication
Scalar multiplication involves multiplying a vector by a scalar, which is a single real number from the field. This operation changes the magnitude of the vector without affecting its direction. For example, if you have a vector \( \mathbf{v} \) and a scalar \( k \), scalar multiplication is represented as \( k \mathbf{v} \).
  • One important property of scalar multiplication is that it distributes over vector addition: \( k(\mathbf{u} + \mathbf{v}) = k\mathbf{u} + k\mathbf{v} \).
  • Scalar multiplication is also associative with respect to the scalar values: \( (cd)\mathbf{v} = c(d\mathbf{v}) \).
These rules allow scalar multiplication to adjust the scale of a vector, thereby offering flexibility to manipulate vectors to have desired lengths or directions. It also ensures that vectors can be scaled consistently across operations.
Additive Inverse
The additive inverse is a property that states for every vector \( \mathbf{v} \) in a vector space, there exists another vector \( -\mathbf{v} \) such that their sum equals the zero vector: \( \mathbf{v} + (-\mathbf{v}) = \mathbf{0} \). This concept ensures that every vector has a counterpart that can "cancel" it out.
  • The additivity of this inverse makes it crucial for simplifying expressions and solving vector equations.
  • The identity \( \mathbf{v} + \mathbf{0} = \mathbf{v} \) is always maintained where \( \mathbf{0} \) is the zero vector.
The existence of additive inverses helps in providing solutions to vector equations by allowing vectors to be negated effectively, maintaining the balance and properties necessary to define this space.
Zero Vector
The zero vector is a special vector in a vector space, denoted often as \( \mathbf{0} \). It plays a crucial role as it acts as the identity element for addition. No matter which vector it is added to, the result is always that vector: \( \mathbf{v} + \mathbf{0} = \mathbf{v} \).
  • The zero vector also satisfies the condition \( 0\mathbf{v} = \mathbf{0} \) for any vector \( \mathbf{v} \), representing the scalar multiplication effect.
  • In any vector space, we also find that adding the additive inverse \( -\mathbf{v} \) results in the zero vector again \( \mathbf{v} + (-\mathbf{v}) = \mathbf{0} \).
Thus, the zero vector is essential in structuring vector space as it supports the underlying arithmetic operations, like addition and scalar multiplication, that define the nature of the vector space.

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Most popular questions from this chapter

Let \(W\) be the subset of continuous functions on [0,1] such that \(f(0)=0\). Prove that \(W\) is a subspace of \(C[0,1]\)

Prove that any set of vectors containing \(\mathbf{0}\) is linearly dependent.

If \(F\) is a field, show that \(F[x]\) is a vector space over \(F\), where the vectors in \(F[x]\) are polynomials. Vector addition is polynomial addition, and scalar multiplication is defined by \(\alpha p(x)\) for \(\alpha \in F\).

Which of the following sets are subspaces of \(\mathbb{R}^{3}\) ? If the set is indeed a subspace, find a basis for the subspace and compute its dimension. (a) \(\left\\{\left(x_{1}, x_{2}, x_{3}\right): 3 x_{1}-2 x_{2}+x_{3}=0\right\\}\) (b) \(\left\\{\left(x_{1}, x_{2}, x_{3}\right): 3 x_{1}+4 x_{3}=0,2 x_{1}-x_{2}+x_{3}=0\right\\}\) (c) \(\left\\{\left(x_{1}, x_{2}, x_{3}\right): x_{1}-2 x_{2}+2 x_{3}=2\right\\}\) (d) \(\left\\{\left(x_{1}, x_{2}, x_{3}\right): 3 x_{1}-2 x_{2}^{2}=0\right\\}\)

Direct Sums. \(\quad\) Let \(U\) and \(V\) be subspaces of a vector space \(W\). The sum of \(U\) and \(V,\) denoted \(U+V\), is defined to be the set of all vectors of the form \(u+v,\) where \(u \in U\) and \(v \in V\) (a) Prove that \(U+V\) and \(U \cap V\) are subspaces of \(W\). (b) If \(U+V=W\) and \(U \cap V=\mathbf{0},\) then \(W\) is said to be the direct sum. In this case, we write \(W=U \oplus V\). Show that every element \(w \in W\) can be written uniquely as \(w=u+v,\) where \(u \in U\) and \(v \in V\) (c) Let \(U\) be a subspace of dimension \(k\) of a vector space \(W\) of dimension \(n\). Prove that there exists a subspace \(V\) of dimension \(n-k\) such that \(W=U \oplus V\). Is the subspace \(V\) unique? (d) If \(U\) and \(V\) are arbitrary subspaces of a vector space \(W\), show that $$ \operatorname{dim}(U+V)=\operatorname{dim} U+\operatorname{dim} V-\operatorname{dim}(U \cap V) $$
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