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Chapter 3: Problem 9

Show that a vector \(v \in V\) is zero if and only if \(f(v)=0\) for all \(f \in V^{*}\).

Short Answer

Expert verified
To prove that a vector \(v \in V\) is zero if and only if \(f(v)=0\) for all \(f \in V^{*}\), we need to prove both directions of the equivalence: 1. If \(f(v)=0\) for all \(f \in V^{*}\), then \(v = 0\). Assuming \(v\neq 0\), we can find a linear transformation \(f^{'}\) such that \(f^{'}(v)\neq 0\), which contradicts our assumption. Hence, \(v = 0\). 2. If \(v=0\), then \(f(v) = 0\) for all \(f \in V^{*}\). This follows directly from the fact that \(f(0) = 0\). Thus, a vector \(v\) in \(V\) is zero if and only if \(f(v)=0\) for all \(f\) in \(V^{*}\).

Step by step solution

01

Understanding the Definitions

A dual space, denoted \(V^{*}\), is the set of all linear transformations that take a vector from \(V\) to the field \(\mathbb{F}\). A linear transformation is a function \(f: V \to \mathbb{F}\) that has two properties: Additivity, \(f(v + w) = f(v) + f(w)\), and homogeneity, \(f(cv) = cf(v)\).
02

Proving the 'If' Part

The 'if' part of the theorem states that if \(f(v)=0\) for all \(f \in V^{*}\), then \(v = 0\). Let's assume \(v\neq 0\). For a non-zero vector, we can find a linear transformation \(f^{'}\) in \(V^{*}\) such that \(f^{'}(v)\neq 0\). This contradicts with our assumption that \(f(v)=0\) for all \(f\in V^{*}\). Hence, our assumption that \(v\neq 0\) must be false. Therefore, if \(f(v) = 0\) for all \(f\in V^{*}\), then \(v\) must equal to zero.
03

Proving the 'Only If' Part

The 'only if' part states that if \(v = 0\), then \(f(v) = 0\) for all \(f \in V^{*}\). This can be proved directly from the properties of a linear transformation: \(f(0) = 0\). Therefore, it holds that \(f(v) = 0\) for all \(f\in V^{*}\) when \(v=0\).
04

Conclusion

By proving both the 'if' and 'only if' components of this theorem, we have shown that a vector \(v\) in \(V\) is zero if and only if \(f(v)=0\) for all \(f\) in \(V^{*}\). This result allows us to tell whether a vector \(v\) in \(V\) is the zero vector by checking the output of all \(f\) in \(V^{*}\).

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Key Concepts

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

Linear Transformation
Linear transformations are one of the cornerstones of algebra, serving as a bridge connecting multiple vector spaces. Essentially, a linear transformation is a rule, denoted by a function f, that maps elements (vectors) from one vector space V to another space or field ℓ. For a transformation to be linear, it must obey two critical rules: additivity and homogeneity.

Additivity means that if you have two vectors, v and w, in your vector space, then f(v + w) = f(v) + f(w). In other words, when you apply the transformation to the sum of two vectors, it's the same as if you transformed each vector separately and then added the results.

Homogeneity deals with multiplication by a scalar (a number from the field ℓ). It dictates that f(cv) = cf(v), where c is a scalar and v is a vector. This means that when a vector is stretched or shrunk, the transformation should reflect that change proportionally.

A proper understanding of linear transformations is crucial because it affects how we interpret solutions in various vector space scenarios, such as in our exercise about the zero vector and dual spaces.
Vector Space
Diving deeper into the realm of linear algebra, one will frequently encounter the abstract concept of a vector space. It's a collection of objects known as vectors, which can be added together and multiplied by scalars to produce another vector. To qualify as a vector space, the set must satisfy certain axioms pertaining to addition and scalar multiplication.

For instance, consider vector addition; it must be associative and commutative, there must be a zero vector that acts as an identity element for addition, and each vector should have an additive inverse. Scalar multiplication needs to behave properly as well, indicating that multiplying any vector by the number 1 results in the vector itself.

Vector spaces are not restricted to any particular dimension and can range from simple 2-dimensional planes to complex infinite-dimensional function spaces. Understanding the structure of vector spaces is fundamental when discussing linear transformations, as emphasized by the dual space interconnectedness in our exercise.
Zero Vector
At the heart of every vector space lies a unique element known as the zero vector. Denoted commonly by 0, the zero vector is the additive identity in vector space; it adds no value when combined with any other vector. Mathematically, for any vector v in vector space V, we have v + 0 = v. It is the equivalent of the number zero in arithmetic but extended to the dimensionality of vectors.

The zero vector plays a pivotal role in understanding vector spaces and linear transformations. In the context of our exercise, any linear transformation, when applied to the zero vector, will always result in the scalar zero. This principle demonstrates the consistency of linear transformations when dealing with base elements of vector spaces.

Keep in mind that the zero vector's properties are essential when analyzing the behavior of entire sets of vectors under transformations, which is evident in the relationship explored between vectors, their dual space, and linear transformations.

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Most popular questions from this chapter

Show that for any nonzero vector \(v \in V\), there exists a linear functional \(f \in V^{*}\) for which \(f(v) \neq 0\).

If \(V\) is infinite-dimensional and \(S\) is an infinite-dimensional subspace, must the dimension of \(V / S\) be finite? Explain.

Let \(S\) be a proper subspace of a finite-dimensional vector space \(V\) and let \(v \in V \backslash S\). Show that there is a linear functional \(f \in V^{*}\) for which \(f(v)=1\) and \(f(s)=0\) for all \(s \in S\).

Let \(\tau \in \mathcal{L}(V)\) and suppose that \(S\) is a subspace of \(V\). Define a map \(\tau^{\prime}: V / S \rightarrow V / S\) by $$ \tau^{\prime}(v+S)=\tau v+S $$ When is \(\tau^{\prime}\) well-defined? If \(\tau^{\prime}\) is well-defined, is it a linear transformation? What are \(\operatorname{im}\left(\tau^{\prime}\right)\) and \(\operatorname{ker}\left(\tau^{\prime}\right)\) ?

Prove that \(0^{\times}=0\) and \(\iota^{\times}=\iota\) where 0 is the zero linear operator and \(\iota\) is the identity.
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