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Chapter 2: Problem 12

(Hyperplane) If \(Y\) is a subspace of a vector space \(X\) and \(\operatorname{codim} Y=1\) (cf. Sec. 2.1, Prob. 14), then every element of \(X / Y\) is called a hyperplane parallel to \(Y\). Show that for any linear functional \(f \neq 0\) on \(X\), the set \(H_{1}=\\{x \in X \mid f(x)=1\\}\) is a hyperplane parallel to the null space \(\mathcal{N}(f)\) of \(f\)

Short Answer

Expert verified
The set \( H_1 \) is a hyperplane parallel to the null space \( \mathcal{N}(f) \).

Step by step solution

01

Define Codimension

The codimension of a subspace \( Y \) in a vector space \( X \) is defined as \( \operatorname{codim} Y = \dim X - \dim Y \). Given that \( \operatorname{codim} Y = 1 \), it means \( Y \) is co-dimension 1 in \( X \), which implies that \( Y \) is just one dimension short of \( X \).
02

Understand Hyperplane Definition

A hyperplane in a vector space is a subspace whose dimension is one less than the dimension of the whole space. Since \( \operatorname{codim} Y = 1 \), \( Y \) itself is a hyperplane in \( X \).
03

Define Linear Functional Null Space

For a nonzero linear functional \( f \) on \( X \), the null space, \( \mathcal{N}(f) = \{ x \in X \mid f(x) = 0 \} \), is a subspace of \( X \). It consists of all vectors that are mapped to zero by \( f \).
04

Analyze \( H_1 \) Set

The set \( H_1 = \{ x \in X \mid f(x) = 1 \} \) is not empty because \( f eq 0 \), ensuring that there must be some vector in \( X \) such that \( f(x) = 1 \). This set represents a collection of vectors which are displaced from \( \mathcal{N}(f) \) by a constant amount (i.e., \( 1 \)).
05

Show \( H_1 \) is Parallel to \( \mathcal{N}(f) \)

Every vector \( x \in H_1 \) can be expressed as \( x = n + v \) where \( n \in \mathcal{N}(f) \) and \( v \) is a vector such that \( f(n + v) = f(n) + f(v) = 1 \). This shows that \( H_1 \) is formed by translating each element in \( \mathcal{N}(f) \) by the vector \( v \) where \( f(v) = 1 \), hence \( H_1 \) is parallel to \( \mathcal{N}(f) \).
06

Conclude that \( H_1 \) is a Hyperplane

Since \( H_1 \) is parallel to \( \mathcal{N}(f) \) and our translation preserves the properties of having one less dimension than \( X \), \( H_1 \) is a hyperplane relative to \( \mathcal{N}(f) \), itself a hyperplane within \( X \). Thus, \( H_1 \) is indeed a hyperplane parallel to \( \mathcal{N}(f) \).

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Key Concepts

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

Codimension
Codimension is an essential concept when studying subspaces and hyperplanes. It is defined as the difference between the dimension of a vector space, say \( X \), and the dimension of a subspace \( Y \) of \( X \). Mathematically, it's expressed as \( \text{codim} \, Y = \dim X - \dim Y \). Codimension gives us a sense of how large a subspace is relative to its ambient space.

  • A subspace \( Y \) with \( \text{codim} \, Y = 1 \) is just one dimension short of \( X \).
  • This implies that \( Y \) divides \( X \) into two parts, forming a hyperplane, which is itself a subspace of \( X \).
Understanding codimension helps in identifying these thin slices within the vector space, which is crucial in fields like linear algebra and geometry.
Linear Functional
A linear functional is a specific type of linear transformation that maps a vector space \( X \) to its field of scalars, usually \( \mathbb{R} \) or \( \mathbb{C} \). Basically, it takes vectors and assigns real or complex numbers to them.

  • A linear functional \( f: X \to \mathbb{R} \) (or \( \mathbb{C} \)) has the property of linearity: \( f(ax + by) = af(x) + bf(y) \) where \( a \) and \( b \) are scalars and \( x, y \in X \).
This type of transformation is essential because it simplifies problems by reducing dimensions. They are used in defining concepts like dual spaces, where each vector space has an associated space formed by its linear functionals.
Null Space
The null space of a linear functional \( f \), denoted \( \mathcal{N}(f) \), is a set of all vectors in \( X \) that are mapped to zero by \( f \). In other words, \( \mathcal{N}(f) = \{ x \in X \mid f(x) = 0 \} \).

  • The null space is a subspace of \( X \).
  • It essentially tells us the direction along which the linear functional vanishes, providing critical insight into the geometry of the problem.
Knowing the null space is important in solving linear equations and understanding the level sets of functions, especially because it helps establish the parallel nature of hyperplanes like \( H_1 \).
Subspace
A subspace is a vector space that is entirely contained within another vector space. For instance, given a vector space \( X \), a subset \( Y \) of \( X \) is a subspace if it satisfies three conditions: it includes the zero vector, it is closed under addition, and it is closed under scalar multiplication.

  • Hyperplanes are specific types of subspaces which are one dimension shy of the total space \( X \).
  • In the context of the exercise, the null space \( \mathcal{N}(f) \) is a subspace of \( X \).
Subspaces serve as foundational structures in linear algebra. They help in the decomposition of spaces and in understanding the solution sets of linear systems.

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Most popular questions from this chapter

If \(f\) is a linear functional on an \(n\)-dimensional vector space \(X\), what dimension can the null space \(\mathcal{N}(f)\) have?

If \(M\) is a linearly dependent set in a complex vector space \(X\), is \(M\) linearly dependent in \(X\), regarded as a real vector space?

If \(x\) and \(y\) are different vectors in a finite dimensional vector space \(X\), show that there is a linear functional \(f\) on \(X\) such that \(f(x) \neq f(y)\).

Show that the functionals defined on \(C[a, b]\) by $$ \begin{aligned} &f_{1}(x)=\int_{a}^{b} x(t) y_{0}(t) d t \\ &f_{2}(x)=\alpha x(a)+\beta x(b) \end{aligned} $$ are linear and bounded.

If \(T \neq 0\) is a bounded linear operator, show that for any \(x \in \mathscr{I}(T)\) such that \(\|x\|<1\) we have the strict inequality \(\|T x\|<\|T\|\).
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