/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 Gegeben sei ein endlichdimension... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 1

Gegeben sei ein endlichdimensionaler Vektorraum \(V\) mit Basen \(\mathcal{A}\) und \(\mathcal{B}\). Sind \(\mathcal{A}^{*}\) und \(\mathcal{B}^{*}\) die zugeh枚rigen dualen Basen von \(V^{*}\), so gilt f眉r die Transformationsmatrizen $$ T_{\mathcal{B}^{*}}^{\mathcal{A}^{*}}=\left({ }^{t} T_{\mathcal{B}}^{\mathcal{A}}\right)^{-1}. $$

Short Answer

Expert verified
The dual transformation matrix is the inverse of the transpose of the original transformation matrix.

Step by step solution

01

Understanding the Problem

We need to prove a relationship between the transformation matrices of dual bases of a vector space. We have two bases \( \mathcal{A} \) and \( \mathcal{B} \) in a vector space \( V \), and their dual bases \( \mathcal{A}^{*} \) and \( \mathcal{B}^{*} \) in the dual space \( V^{*} \). The goal is to show \( T_{\mathcal{B}^{*}}^{\mathcal{A}^{*}} = \left( {}^{t} T_{\mathcal{B}}^{\mathcal{A}} \right)^{-1} \).
02

Definition and Properties of Dual Bases

The dual basis \( \mathcal{A}^{*} \) consists of linear functionals that satisfy \( \mathcal{A}^{*}(\mathcal{A}) = I \), where \( I \) is the identity matrix. Each functional in \( \mathcal{A}^{*} \) maps vectors in \( \mathcal{A} \) to their components. Similarly, \( \mathcal{B}^{*}(\mathcal{B}) = I \). The transformation matrix \( T_{\mathcal{B}}^{\mathcal{A}} \) converts coordinates from basis \( \mathcal{B} \) to \( \mathcal{A} \).
03

Transpose and Inverse Relationship

If \( T_{\mathcal{B}}^{\mathcal{A}} \) is the matrix representing the change of basis from \( \mathcal{B} \) to \( \mathcal{A} \), the transpose \( {}^t T_{\mathcal{B}}^{\mathcal{A}} \) represents the map from the dual basis \( \mathcal{A}^{*} \) to \( \mathcal{B}^{*} \). The inverse \( \left( {}^t T_{\mathcal{B}}^{\mathcal{A}} \right)^{-1} \) then gives the transformation matrix from \( \mathcal{B}^{*} \) to \( \mathcal{A}^{*} \).
04

Prove the Dual Transformation Relationship

The map from dual basis \( \mathcal{A}^{*} \) to \( \mathcal{B}^{*} \) is the transpose \( {}^t T_{\mathcal{B}}^{\mathcal{A}} \). Therefore, the transformation from \( \mathcal{B}^{*} \) to \( \mathcal{A}^{*} \) is its inverse \( \left( {}^t T_{\mathcal{B}}^{\mathcal{A}} \right)^{-1} \). This is a consequence of the fact that in the dual space, operations are represented by transposes of the original operations. Hence, \( T_{\mathcal{B}^{*}}^{\mathcal{A}^{*}} = \left( {}^{t} T_{\mathcal{B}}^{\mathcal{A}} \right)^{-1} \).
05

Conclusion and Verification

Then the expression \( T_{\mathcal{B}^{*}}^{\mathcal{A}^{*}} = \left( {}^{t} T_{\mathcal{B}}^{\mathcal{A}} \right)^{-1} \) is verified using the properties of dual spaces and their operations on the transformation matrices. This concludes the proof.

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Key Concepts

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

Vector Space
A vector space is an essential building block in various fields of mathematics, especially linear algebra. If you're new to the concept, think of it as a collection of objects called vectors, which can be added together and multiplied by numbers, called scalars, to gain new vectors. To break it down:
  • A vector is any entity that has direction and magnitude.
  • Scalars are numerical values that scale or resize vectors without changing their direction.
  • Common operations in vector spaces include vector addition and scalar multiplication.
Vector spaces need to satisfy specific properties, such as closure under addition and scalar multiplication, the existence of an additive identity (a zero vector), and more. It鈥檚 the structure that allows us to analyze spaces like the 2D plane or higher-dimensional analogs. Understanding vector spaces provides a foundation to explore more complex topics, such as dual bases and the transformations between different bases.
Transformation Matrix
A transformation matrix is a fundamental tool in linear algebra, representing linear transformations from one vector space basis to another. In simple terms, it enables you to express a vector in a new coordinate system. Here's how to understand it better:
  • Imagine you have two different ways to describe vectors in a vector space, called Basis A and Basis B.
  • A transformation matrix allows you to convert vector representations from one basis (like Basis B) to another (like Basis A).
  • The transformation involves matrix multiplication: coordinate change is expressed as a product of the vector representation in the initial basis with the transformation matrix.
This matrix is crucial in areas requiring basis changes, such as computer graphics, where it helps rotate and scale drawings efficiently. The relationship between the transformation matrices of dual bases, as mentioned in the exercise, involves the transpose and inverse, linking how dual vectors transform relative to their original bases.
Transpose and Inverse
The concepts of transpose and inverse are pivotal in understanding matrix operations. Each has a unique role, especially within the context of dual bases in vector spaces.
  • The transpose of a matrix is achieved by flipping it over its diagonal, which means the top-right element becomes the bottom-left one, and so forth. In mathematical notation, for a matrix \( A \), its transpose is denoted as \( {}^{t}A \).
  • The inverse of a matrix is a matrix that, when multiplied with the original, yields the identity matrix. Not all matrices have inverses; a matrix must be square and have a non-zero determinant to have an inverse. The inverse of a matrix \( A \) is denoted as \( A^{-1} \).
  • In the given exercise, the inverse of the transpose \( {}^{t} T_{B}^{A} \) describes how dual spaces operate. Specifically, it provides a transformation from one dual basis to another, showing their intrinsic relationship and influence on dual vector spaces.
Together, the transpose and inverse connect the dots between transformations in standard spaces and their duals, establishing crucial relationships seen in dual basis transformations.

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Most popular questions from this chapter

Es sei \(V\) ein Vektorraum 眉ber einen K枚rper \(K\) und \(L \supset K\) ein Erweiterungsk枚rper von \(L\), d.h. \(L\) ist ein K枚rper und \(K\) ein Unterring von \(L\) (vgl. 1.3.2). a) Zeigen Sie, dass \(L\) eine Struktur als \(K\)-Vektorraum tr盲gt. b) F眉r Elemente \(\sum \lambda_{i} \otimes v_{i} \in L \otimes_{K} V\) und \(\lambda \in L\) definieren wir eine skalare Multiplikation durch $$ \lambda \cdot\left(\sum \lambda_{i} \otimes v_{i}\right):=\sum \lambda \lambda_{i} \otimes v_{i} $$ Zeigen Sie, dass \(L \otimes_{K} V\) mit der 眉blichen Addition und dieser skalaren Multiplikation zu einem \(L\)-Vektorraum wird. c) Ist die Familie \(\left(v_{i}\right)_{i \in I}\) eine Basis von \(V\) 眉ber \(K\), so ist die Familie \(\left(1 \otimes v_{i}\right)_{i \in I}\) eine Basis von \(L \otimes_{K} V\) 眉ber \(L\). Insbesondere gilt \(\operatorname{dim}_{K} V=\operatorname{dim}_{L}\left(L \otimes_{K} V\right)\). d) Durch die Abbildung $$ \varphi: V \rightarrow K \otimes_{K} V, \quad v \mapsto 1 \otimes v $$ ist ein Isomorphismus von \(K\)-Vektorr盲umen gegeben.

Gegeben seien zwei windschiefe Geraden \(L=v+\mathbb{R} w\) und \(L^{\prime}=v^{\prime}+\mathbf{R} w^{\prime} \operatorname{im} \mathbb{R}^{n}\). Wir wollen zwei Methoden angeben, um den Abstand $$ d\left(L, L^{\prime}\right)=\min \left\\{d\left(u, u^{\prime}\right)=\left\|u^{\prime}-u\right\|: u \in L, u^{\prime} \in L^{\prime}\right\\} $$ zu berechnen. Zur Vereinfachung nehmen wir \(\|w\|=\left\|w^{\prime}\right\|=1\) an und definieren $$ \delta: \mathbb{R}^{2} \rightarrow \mathbb{R}, \quad\left(\lambda, \lambda^{\prime}\right) \mapsto\left\|v^{\prime}+\lambda^{\prime} w^{\prime}-v-\lambda w\right\|^{2} $$ a) Untersuchen Sie die Funktion \(\delta\) mit Hilfe der Differentialrechnung auf Extrema und bestimmen damit den Abstand \(d\left(L, L^{\prime}\right)\). b) Es gilt \(\delta\left(\lambda, \lambda^{\prime}\right)=\lambda^{2}+a \lambda \lambda^{\prime}+\lambda^{\prime 2}+b \lambda+c \lambda^{\prime}+d .\) Setzen Sie \(\mu:=\lambda+\frac{a}{2} \lambda^{\prime}\) und \(\mu^{\prime}=\frac{\sqrt{4-a^{2}}}{2} \lambda^{\prime}\) und zeigen Sie, dass man auf diese Weise \(\delta\) durch quadratische Erg盲nzung schreiben kann als \(\delta\left(\lambda, \lambda^{\prime}\right)=(\mu-e)^{2}+\left(\mu^{\prime}-f\right)^{2}+g .\) Dann ist \(g=d\left(L, L^{\prime}\right)\).

F眉r \(K\)-Vektorr盲ume \(V, V^{\prime}, W, W^{\prime}\) sowie Homomorphismen \(F: \quad V \rightarrow V^{\prime}\) und \(G: W \rightarrow W^{\prime}\) definieren wir das Tensorprodukt von \(F\) und \(G\) durch $$ \begin{aligned} (F \otimes G): V \otimes W & \rightarrow \quad V^{\prime} \otimes W^{\prime}, \\\ v \otimes w & \mapsto F(v) \otimes G(w). \end{aligned} $$ Zeigen Sie, dass hierdurch ein Vektorraum-Isomorphismus $$ \operatorname{Hom}_{K}\left(V, V^{\prime}\right) \otimes \operatorname{Hom}_{K}\left(W, W^{\prime}\right) \rightarrow \operatorname{Hom}_{K}\left(V \otimes W, V^{\prime} \otimes W^{\prime}\right) $$ definiert wird.

Sei \(A \in \mathrm{M}(n \times n ; \mathbb{C})\) antihermitesch, das hei脽t \(-A={ }^{t} \bar{A}\). Zeigen Sie, dass \(A\) normal ist und alle Eigenwerte von \(A\) in iR liegen.

\(V\) sei ein endlichdimensionaler \(K\)-Vektorraum und \(\alpha=\left(\alpha_{1} \wedge \ldots \wedge \alpha_{k}\right) \in \Lambda^{k} V\) sowie \(\beta=\left(\beta_{1} \wedge \ldots \wedge \beta_{l}\right) \in \bigwedge^{l} V\) a) Zeigen Sie, dass eine bilineare Abbildung $$ \mu: \bigwedge^{k} V \times \Lambda^{l} V \rightarrow \Lambda^{k+l} V $$ mit $$ (\alpha, \beta) \mapsto \alpha_{1} \wedge \ldots \wedge \alpha_{k} \wedge \beta_{1} \wedge \ldots \wedge \beta_{l} $$ existiert. Das Element \(\alpha \wedge \beta:=\mu(\alpha, \beta)\) hei脽t 盲u\betaeres Produkt von \(\alpha\) und \(\beta\). b) Es gilt $$ \alpha \wedge \beta=(-1)^{k \cdot l} \beta \wedge \alpha. $$
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