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Chapter 5: Problem 2

Sei \(\mathcal{D}=\mathcal{D}(]-1,1[; \mathbb{R})\) der Vektorraum der auf ] \(-1,1\) [ differenzierbaren Funktionen. a) Zeigen Sie, dass \(d: \mathcal{D} \times \mathcal{D} \rightarrow \mathbf{R},(f, g) \mapsto(f g)^{\prime}(0)\) eine symmetrische Bilinearform ist. b) Bestimmen Sie den Ausartungsraum \(\mathcal{D}_{0}\) von \(d\).

Short Answer

Expert verified
\(d(f, g)\) is symmetric bilinear; \(\mathcal{D}_0\) is functions with \(f'(0) = 0\), \(f(0) = 0\).

Step by step solution

01

Understanding the Problem

We need to show that the map \(d: \mathcal{D} \times \mathcal{D} \rightarrow \mathbb{R}\), defined by \(d(f, g) = (f g)'(0)\), is a symmetric bilinear form. Additionally, we must determine the degeneracy space, \(\mathcal{D}_0\), of \(d\).
02

Checking Bilinearity

A function is bilinear if it is linear in each argument separately. Thus, we first fix \(g\) and check linearity in \(f\):d(f+f', g) = [(f+f')g]'(0) = (fg + f'g)'(0) = (fg)'(0) + (f'g)'(0) = d(f,g) + d(f',g).Similarly, scalar multiplication gives:d(cf, g) = [cfg]'(0) = c(fg)'(0) = c \cdot d(f,g).Next, fix \(f\) and check linearity in \(g\):d(f, g+g') = [f(g+g')]'(0) = (fg+fg')'(0) = (fg)'(0) + (fg')'(0) = d(f,g) + d(f,g').And,d(f, cg) = [fcg]'(0) = c(fg)'(0) = c \cdot d(f,g).Therefore, \(d\) is linear in both variables, confirming bilinearity.
03

Checking Symmetry

A function is symmetric if \(d(f, g) = d(g, f)\) for all \(f, g \in \mathcal{D}\). We calculate:(d(f, g) = (fg)'(0) = f'(0)g(0) + f(0)g'(0),and,(d(g, f) = (gf)'(0) = g'(0)f(0) + g(0)f'(0).Since the expressions are equal, \(d(f, g) = d(g, f)\). Thus, \(d\) is symmetric.
04

Finding the Degeneracy Space

The degeneracy space \(\mathcal{D}_0\) consists of all \(f \in \mathcal{D}\) such that \(d(f, g) = 0\) for all \(g \in \mathcal{D}\). Thus, we need \((fg)'(0) = 0\) for every \(g\). This implies \(f'(0)g(0) + f(0)g'(0) = 0\). Since this must hold for all \(g\), consider \(g(x) = 1\), leading to \(f'(0) = 0\), and \(g(x) = x\), giving \(f(0) = 0\). So the degeneracy space \(\mathcal{D}_0\) consists of functions with \(f'(0) = 0\) and \(f(0) = 0\).

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Key Concepts

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

Degeneracy Space
The concept of a degeneracy space, often denoted as \(\mathcal{D}_0\), is vital in understanding bilinear forms. A bilinear form can map two inputs to a single output, typically a real number. When examining such a form, the degeneracy space consists of all vectors that interact with any vector through the bilinear form and result in zero. In simpler terms, it is the set of functions that "degenerate" the bilinear pairing to zero. Let's see how it applies here:
  • The degeneracy space \(\mathcal{D}_0\) is characterized by those functions \(f\) such that the bilinear form \(d(f, g)\) equals zero for every function \(g\) in the space.
  • With the specific linear form \(d(f, g) = (fg)'(0)\), we need \((fg)'(0) = 0\) consistently for all possible \(g\).
  • This occurs when \(f'(0) = 0\) and \(f(0) = 0\), meaning both the function and its derivative at zero must be zero for it to belong to the degeneracy space.
Understanding these criteria helps identify all such candidate functions, simplifying the characterization of \(\mathcal{D}_0\) in complex scenarios.
Symmetric Form
A symmetric form simply means that the order of the inputs does not affect the result. In more formal terms, a function \(d(f, g)\) is symmetric if it holds true that \(d(f, g) = d(g, f)\) for all involved functions \(f\) and \(g\). This inherent property makes symmetric forms quite valuable in mathematics:
  • For the given problem, we need to establish that \(d(f, g) = (fg)'(0) = f'(0)g(0) + f(0)g'(0)\) is equal to \(d(g, f) = (gf)'(0) = g'(0)f(0) + g(0)f'(0)\).
  • Evaluating both expressions shows that they are indeed equal, confirming the symmetric nature of the bilinear form \(d(f, g)\).
  • The symmetric property is quite intuitive here because the multiplication within the derivative allows for interchangeability of \(f\) and \(g\), thus preserving symmetry.
This interchangeability often simplifies mathematical proofs and problem-solving strategies.
Vector Space
A vector space is a fundamental concept in linear algebra, consisting of a set of vectors where vector addition and scalar multiplication are defined and satisfy certain properties. Functions defined over certain intervals form vector spaces when they can be added together and multiplied by scalars while staying within the defined space.
  • The vector space \(\mathcal{D}\) in this scenario is composed of differentiable functions over the interval \((-1,1)\).
  • Within this space, considering any two functions \(f\) and \(g\), their sum \(f+g\) and any scalar multiple \(cf\) of a function \(f\) also remain differentiable over the same interval.
  • These operations satisfy the vector space properties: closure, associativity, distributivity, existence of an additive identity, and the existence of additive inverses for each vector.
Recognizing \(\mathcal{D}\) as a vector space permits us to use linear algebra techniques to solve problems and find structures like basis and dimension.
Differentiable Functions
Differentiable functions are those functions that have derivatives at every point in their domain. This means the function behaves smoothly without jumps, sharp points, or cusps, at least over the interval considered:
  • In the context of this exercise, we are dealing with differentiable functions defined on the open interval \((-1, 1)\).
  • A function \(f: (-1, 1) \rightarrow \mathbb{R}\) is differentiable if its derivative \(f'(x)\) exists for every \(x\) in \((-1, 1)\).
  • Such functions are particularly suitable for defining vector spaces and bilinear forms because they maintain continuity and smooth interaction across computations.
A solid understanding of differentiability helps predict the behavior of functions and manage sophisticated operations like those involved in this bilinear form exercise.

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Most popular questions from this chapter

Seien \(F\) und \(G\) zwei selbstadjungierte Endomorphismen auf einem endlichdimensionalen euklidischen bzw. unit盲ren Vektorraum \(V\). Zeigen Sie, dass \(F \circ G\) selbstadjungiert ist genau dann, wenn \(F \circ G=G \circ F\).

Sei \(V\) ein 3-dimensionaler \(\mathbb{R}\)-Vektorraum, \(\mathcal{A}=\left(v_{1}, v_{2}, v_{3}\right)\) eine Basis von \(V\) und \(s\) eine Bilinearform auf \(V\) mit $$ M_{\mathcal{A}}(s)=\left(\begin{array}{lll} 1 & 1 & 2 \\ 1 & 1 & 1 \\ 0 & 1 & 1 \end{array}\right) $$ Zeigen Sie, dass \(\mathcal{B}=\left(v_{1}+v_{2}, v_{2}+v_{3}, v_{2}\right)\) eine Basis von \(V\) ist, und berechnen Sie \(M_{\mathcal{B}}(s)\)

Wir wollen mit Hilfe des Vektorproduktes eine Parameterdarstellung der Schnittgeraden zweier nichtparalleler Ebenen im \(\mathbb{R}^{3}\) bestimmen. Sind zwei Ebenen \(E=v+\mathbb{R} w_{1}+\mathbf{R} w_{2}, E^{\prime}=v^{\prime}+\mathbf{R} w_{1}^{\prime}+\mathbf{R} w_{2}^{\prime} \subset \mathbb{R}^{3}\) gegeben, so sei \(W=\mathbf{R} w_{1}+\mathbf{R} w_{2}\) \(W^{\prime}=\mathbf{R} w_{1}^{\prime}+\mathbf{R} w_{2}^{\prime}\). Da die beiden Ebenen nicht parallel sind, ist \(W \neq W^{\prime}\), und damit hat \(U=W \cap W^{\prime}\) die Dimension 1. Zeigen Sie: a) Ist \(L=E \cap E^{\prime}\) und \(u \in L\), so ist \(L=u+U\). b) Seien \(s=w_{1} \times w_{2}, s^{\prime}=w_{1}^{\prime} \times w_{2}^{\prime}\) und \(w=s \times s^{\prime}\). Dann gilt \(U=\mathbf{R} w\). Bestimmen Sie nach diesem Verfahren eine Parameterdarstellung von \(E \cap E^{\prime}\), wobei $$ \begin{aligned} &E=(0,2,3)+\mathbb{R}(3,6,5)+\mathbb{R}(1,7,-1) \\ &E^{\prime}=(-1,3,2)+\mathbb{R}(8,2,3)+\mathbf{R}(2,-1,-2) \end{aligned} $$

Zeigen Sie, dass f眉r \(F \in O(3)\) gilt: \(F(x) \times F(y)=(\operatorname{det} F) \cdot F(x \times y)\).

Bestimmen Sie f眉r die Matrix$$ A=\left(\begin{array}{rrr} 2 & -1 & 1 \\ -1 & 2 & 1 \\ 1 & 1 & 2 \end{array}\right) $$ eine orthogonale Matrix \(S \in O(3)\), so dass \({ }^{t} S A S\) eine Diagonalmatrix ist.
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