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Chapter 9: Problem 9

Use Zorn's lemma to show that any nontrivial inner product space has a Hilbert basis.

Short Answer

Expert verified
To show that any nontrivial inner product space has a Hilbert basis, first consider the set "S" containing all linearly independent subsets of the inner product space and define the subset order relation 鈯� as the order relation on "S". Since the empty set is linearly independent, "S" is nonempty and partially ordered. Now, take a chain "C" in "S" and consider the union of all sets in "C", denoted as U(C). Show that U(C) is linearly independent, which implies that "C" has an upper bound in "S". Apply Zorn's lemma to conclude that there must be at least one maximal element in "S". Finally, show that the maximal element is a Hilbert basis for the inner product space.

Step by step solution

01

Define the terms

Before we begin, let's define some key terms: - Inner product space: A vector space equipped with an inner product, a function that assigns a scalar to two vectors. - Hilbert basis: A set of vectors whose linear combinations approximate any given vector in the space arbitrarily closely. - Zorn's lemma: In a nonempty partially ordered set (poset) in which every chain (totally ordered subset) has an upper bound, there exists at least one maximal element. Now we will prove that any nontrivial inner product space has a Hilbert basis using Zorn's lemma.
02

Define the poset and the order relation

Let's consider the set of all linearly independent subsets of our inner product space "S". We need to prove that "S" is a nonempty partially ordered set (poset). For this, we need to define an order relation on the elements of "S". We can use the 鈯� (subset) order relation on the linearly independent sets. According to this relation, given two sets A and B, A 鈯� B if every element of A is also an element of B. To show that "S" is nonempty, note that any vector subspace contains the empty set 鈭�. The empty set is linearly independent because it satisfies the property that no linear combination of vectors in the set can equal to the zero vector. Therefore, "S" is nonempty.
03

Apply Zorn's lemma

Let "C" be a chain in "S" (a totally ordered set of linearly independent subsets of the inner product space). We want to show that "C" has an upper bound. Consider the union of all sets in the chain "C", denoted as U(C). We claim U(C) is an upper bound for "C" (meaning that every set in "C" is a subset of U(C)). Since the elements in "C" are linearly independent sets and "C" is a chain, any two sets A, B 鈭� C must satisfy A 鈯� B or B 鈯� A. Thus, the union U(C) includes all elements from every set in "C". Now we need to prove that U(C) is linearly independent. Suppose there exists a linearly dependent vector in U(C), then it can be written as a linear combination of other vectors in U(C). However, this contradicts the fact that all vectors in the sets of "C" are linearly independent. Thus, U(C) is linearly independent. Since the chain "C" has an upper bound in "S", we can now apply Zorn's lemma, which states that there must be at least one maximal element in "S".
04

Show that the maximal element is a Hilbert basis

Let M be a maximal element in "S". We will show that M constitutes a Hilbert basis for our inner product space. If M spans the entire inner product space, then M is indeed a Hilbert basis. Suppose there exists a vector "v" in the inner product space that is orthogonal to every vector in M, where v 鈮� 0. Note that adding "v" to M creates a larger linearly independent set, since "v" cannot be a linear combination of vectors in M. However, this contradicts the maximality of M in "S". Therefore, there are no such orthogonal vectors "v" outside M. This means that any vector in the inner product space can be approached arbitrarily closely by a linear combination of vectors in M. Consequently, M constitutes a Hilbert basis for the inner product space.

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Key Concepts

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

Inner Product Space
An inner product space is like a vector space with a nice additional feature. This feature is called an "inner product."

The inner product is a special function that takes two vectors and returns a scalar. Imagine it like a machine that takes in two inputs (the vectors) and gives out one output (the scalar). This scalar generally provides meaningful information about those two vectors.
  • When the scalar is zero, it means the vectors are orthogonal, that is, at right angles to each other.
  • If the scalar is positive, it indicates the vectors point in similar directions.
  • A negative scalar shows vectors pointing in opposite directions.
Inner product spaces are important as they provide a framework for defining concepts like length and angles in a vector space. This framework is useful in many areas of mathematics and physics, where understanding the relationship and orientation between vectors is crucial.
Zorn's Lemma
Zorn's lemma might sound a bit tricky at first, but it's a powerful tool in set theory. It is particularly useful for proving the existence of certain elements in a set without having to directly construct them.

The lemma states that in any non-empty partially ordered set (often shortened to "poset"), if every chain (totally ordered subset) has an upper bound, then there is at least one maximal element. Here's a breakdown of what this means:
  • **Non-empty set:** The set in question should contain at least one element.
  • **Partially ordered set:** There is some relationship that orders the elements, but not necessarily in a linear fashion.
  • **Chain:** A subset where every pair of elements is comparable. Think of this as a sort of ordered list within the poset.
  • **Upper bound:** An element that is greater than or equal to every element in the chain.
  • **Maximal element:** An element that is not smaller than any other element in the poset in terms of the ordering relation.
Zorn's lemma is important in areas such as algebra and functional analysis, often used to establish the existence of bases and completing space structures.
Linearly Independent Sets
Linearly independent sets are a key concept in linear algebra. A set of vectors is called linearly independent if no vector in the set can be written as a combination of the others.

This idea can initially be a bit abstract, so let's break it down:
  • **Linear combination:** A way to combine vectors by multiplying them by constants (scalars) and adding the results.
  • **Independence:** No vector in the set is redundant or unnecessary. If we tried to express any one vector as a combination of others, we would fail.
Why is this important? Linearly independent sets are crucial when forming a basis for a vector space. They ensure that every dimension or direction in the space is covered without duplication, allowing for efficient representation of the space. In computational and practical terms, working with linearly independent sets allows us to solve equations and understand vector spaces more effectively.

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Most popular questions from this chapter

If \(V\) and \(W\) are inner product spaces, consider the function on \(V \boxplus W\) defined by $$ \left\langle\left(v_{1}, w_{1}\right),\left(v_{2}, w_{2}\right)\right\rangle=\left\langle v_{1}, v_{2}\right\rangle+\left\langle w_{1}, w_{2}\right\rangle $$ Is this an inner product on \(V \boxplus W\) ?

A normed vector space over \(\mathbb{R}\) or \(\mathbb{C}\) is a vector space (over \(\mathbb{R}\) or \(\mathbb{C}\) ) together with a function \|\|\(: V \rightarrow \mathbb{R}\) for which for all \(u, v \in V\) and scalars \(r\) we have a) \(\|r v\|=|r|\|v\|\) b) \(\|u+v\| \leq\|u\|+\|v\|\) c) \(\|v\|=0\) if and only if \(v=0\) If \(V\) is a real normed space (over \(\mathbb{R}\) ) and if the norm satisfies the parallelogram law $$ \|u+v\|^{2}+\|u-v\|^{2}=2\|u\|^{2}+2\|v\|^{2} $$ prove that the polarization identity $$ \langle u, v\rangle=\frac{1}{4}\left(\|u+v\|^{2}-\|u-v\|^{2}\right) $$ defines an inner product on \(V\). Hint: Evaluate \(8\langle u, x\rangle+8\langle v, x\rangle\) to show that \(\langle u, 2 x\rangle=2\langle u, x\rangle\) and \(\langle u, x\rangle+\langle v, x\rangle=\langle u+v, x\rangle\). Then complete the proof that \(\langle u, r x\rangle=r\langle u, x\rangle\).

Let \(u=\left(r_{1}, \ldots, r_{n}\right)\) and \(v=\left(s_{1}, \ldots, s_{n}\right)\) be in \(\mathbb{R}^{n}\). The Cauchy-Schwarz inequality states that $$ \left|r_{1} s_{1}+\cdots+r_{n} s_{n}\right|^{2} \leq\left(r_{1}^{2}+\cdots+r_{n}^{2}\right)\left(s_{1}^{2}+\cdots+s_{n}^{2}\right) $$ Prove that we can do better: $$ \left(\left|r_{1} s_{1}\right|+\cdots+\left|r_{n} s_{n}\right|\right)^{2} \leq\left(r_{1}^{2}+\cdots+r_{n}^{2}\right)\left(s_{1}^{2}+\cdots+s_{n}^{2}\right) $$

Let \(S\) be a subspace of a finite-dimensional inner product space \(V\). Prove that each coset in \(V / S\) contains exactly one vector that is orthogonal to \(S\).

Let \(f\) be a linear functional on a subspace \(S\) of a finite-dimensional inner product space \(V\). Let \(f(v)=\left\langle v, R_{f}\right\rangle\). Suppose that \(g \in V^{*}\) is an extension of \(f\), that is, \(\left.g\right|_{S}=f\). What is the relationship between the Riesz vectors \(R_{f}\) and \(R_{g}\) ?
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