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Chapter 5: Problem 25

Let \(W\) be a subspace of \(\mathbb{R}^{n}\) and \(\mathbf{v}\) a vector in \(\mathbb{R}^{n}\). Suppose that \(\mathbf{w}\) and \(\mathbf{w}^{\prime}\) are orthogonal vectors with \(\mathbf{w}\) in \(W\) and that \(\mathbf{v}=\mathbf{w}+\mathbf{w}^{\prime} .\) Is it necessarily true that \(\mathbf{w}^{\prime}\) is in \(W^{\perp}\) ? Either prove that it is true or find a counterexample.

Short Answer

Expert verified
No, \( \mathbf{w}^{\prime} \) is not necessarily in \( W^{\perp} \).

Step by step solution

01

Understanding the Problem

We are given a vector \( \mathbf{v} = \mathbf{w} + \mathbf{w}^{\prime} \) where \( \mathbf{w} \) is in the subspace \( W \) of \( \mathbb{R}^n \). We need to determine if \( \mathbf{w}^{\prime} \) is necessarily in the orthogonal complement of \( W \), denoted \( W^{\perp} \).
02

Definition of Orthogonal Complement

Recall that the orthogonal complement \( W^{\perp} \) consists of all vectors in \( \mathbb{R}^n \) that are orthogonal to every vector in \( W \). That is, for any vector \( \mathbf{x} \in W \), \( \mathbf{x} \cdot \mathbf{w}^{\prime} = 0 \) should hold if \( \mathbf{w}^{\prime} \in W^{\perp} \).
03

Analyze Given Conditions

The problem states that \( \mathbf{w} \) and \( \mathbf{w}^{\prime} \) are orthogonal vectors. This means \( \mathbf{w} \cdot \mathbf{w}^{\prime} = 0 \). Since \( \mathbf{w} \in W \), \( \mathbf{w}^{\prime} \) is orthogonal to at least \( \mathbf{w} \).
04

Check Condition for \( W^{\perp} \)

To claim \( \mathbf{w}^{\prime} \in W^{\perp} \), \( \mathbf{w}^{\prime} \) must be orthogonal to every vector in \( W \), not just \( \mathbf{w} \). Thus, if \( W \) contains more vectors than just scalar multiples of \( \mathbf{w} \), we cannot guarantee \( \mathbf{w}^{\prime} \) is orthogonal to all vectors in \( W \).
05

Counterexample

Consider \( \mathbb{R}^2 \) with \( W \) being the x-axis. Let \( \mathbf{v} = (1, 1) \), \( \mathbf{w} = (1, 0) \), and \( \mathbf{w}^{\prime} = (0, 1) \). Here, \( \mathbf{v} = \mathbf{w} + \mathbf{w}^{\prime} \), \( \mathbf{w} \in W \), and \( \mathbf{w} \perp \mathbf{w}^{\prime} \). \( \mathbf{w}^{\prime} eq 0 \) is orthogonal to the x-axis, hence is in \( W^{\perp} \). However, this holds as \( \mathbf{w}^{\prime} \) (y-axis) is orthogonal to \( W \) (x-axis). In general, such an orthogonality condition need not hold for subspaces larger than the span of one vector.
06

Conclusion

The specific construction of this problem satisfies the condition, but it is not necessarily true for general subspaces \( W \) in \( \mathbb{R}^n \). Hence, without further constraints on \( W \), we cannot universally claim \( \mathbf{w}^{\prime} \in W^{\perp} \).

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Key Concepts

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

Subspace
A subspace is a fundamental concept in linear algebra. It refers to a set of vectors in a vector space that, like the vector space itself, is closed under vector addition and scalar multiplication. Simple properties to remember include:
  • Zero vector is always in a subspace.
  • If two vectors are in a subspace, their sum is also in that subspace.
  • Multiplying a vector by any scalar does not take it out of the subspace.
By understanding these properties, it becomes clear why subspaces are crucial in defining structures within a vector space. For instance, in our problem, the subspace \( W \) within \( \mathbb{R}^n \) is used to examine the components of a vector \( \mathbf{v} \). This exploration often helps in understanding how certain vectors relate to each other in a structured way.
Orthogonal Complement
The orthogonal complement of a subspace \( W \) is a set of vectors that are orthogonal to every vector in \( W \). Mathematically, if \( \mathbf{x} \cdot \mathbf{y} = 0 \) for every \( \mathbf{x} \in W \), then \( \mathbf{y} \) is in \( W^{\perp} \). Here are some key points:
  • \( W^{\perp} \) is itself a subspace of \( \mathbb{R}^n \).
  • The intersection of \( W \) and \( W^{\perp} \) contains only the zero vector.
  • Every vector in \( \mathbb{R}^n \) can be decomposed uniquely into two components: one in \( W \) and the other in \( W^{\perp} \).
Understanding \( W^{\perp} \) is pivotal in vector decomposition, as it separates the part of a vector that is within a subspace from the part that is orthogonal to it. This concept plays a significant role in many applications, like solving systems of linear equations and computer graphics.
Vector Decomposition
Vector decomposition is the process of breaking down a vector into components that belong to specific subspaces. In our exercise, the vector \( \mathbf{v} \) is expressed as \( \mathbf{v} = \mathbf{w} + \mathbf{w}^{\prime} \), where \( \mathbf{w} \) is in \( W \). Let's explore this concept:
  • Every vector can be decomposed into two orthogonal components with respect to a subspace \( W \): one in \( W \) and the other in \( W^{\perp} \).
  • The decomposition helps clarify how vectors project onto subspaces and orthogonal complements.
In practice, vector decomposition is often used in data analysis, physics, and engineering to simplify complex vector behaviors. It provides insight into how a vector might behave under linear transformations by breaking it down into manageable subparts.
Counterexample
In mathematics, a counterexample is a specific case which disproves a general statement. In our problem about subspaces and orthogonal complements, we use a counterexample to show that the vector \( \mathbf{w}^{\prime} \) need not necessarily belong to \( W^{\perp} \).
Let's recount the counterexample provided:
  • Consider \( W \) as the x-axis in \( \mathbb{R}^2 \).
  • Take \( \mathbf{v} = (1, 1) \), \( \mathbf{w} = (1, 0) \), \( \mathbf{w}^{\prime} = (0, 1) \).
  • Here, \( \mathbf{w} \perp \mathbf{w}^{\prime} \) satisfied the orthogonal condition but covered only a specific scenario where \( \mathbf{w}^{\prime} \) precisely falls on \( W^{\perp} \).
  • This does not hold for general subspaces where \( W \) could be more than a span of a single vector.
Counterexamples are crucial to understanding the boundaries of mathematical truths, reminding us that assumptions can sometimes limit broader applicability.

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Most popular questions from this chapter

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Let \(\mathcal{B}=\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{n}\right\\}\) be an orthonormal basis for \(\mathbb{R}^{n}\) (a) Prove that, for any x and y in \(\mathbb{R}^{n}\) \\[\begin{array}{r}\mathbf{x} \cdot \mathbf{y}=\left(\mathbf{x} \cdot \mathbf{v}_{1}\right)\left(\mathbf{y} \cdot \mathbf{v}_{1}\right)+\left(\mathbf{x} \cdot \mathbf{v}_{2}\right)\left(\mathbf{y} \cdot \mathbf{v}_{2}\right)+\cdots \\ +\left(\mathbf{x} \cdot \mathbf{v}_{n}\right)\left(\mathbf{y} \cdot \mathbf{v}_{n}\right)\end{array}\\] (b) What does Parseval's Identity imply about the relationship between the dot products \(x \cdot y\) and \([\mathbf{x}]_{B} \cdot[\mathbf{y}]_{B} ?\) (This identity is called Parseval's Identity.)
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