Chapter 14: Problem 5
Let \(G\) be the set of all maps of \(\mathbf{R}^{n}\) into itself of type $$ \sigma_{A, B}: X \mapsto A X+B $$ where \(A\) is an invertible \(n \times n\) matrix, and \(B\) is in \(\mathbf{R}^{n}\). Show that \(G\) is a group. Show that the map $$ \sigma_{A, B} \mapsto A $$ is a homomorphism of \(G\) into the general linear group. Deseribe the kernel.
Short Answer
Step by step solution
Closure
Associativity
Identity
Inverses
Homomorphism
Kernel description
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Linear Transformations
To understand this, consider a transformation \(T: \mathbf{R}^n \rightarrow \mathbf{R}^n\) that takes a vector \(X\) and applies a matrix \(A\). Formally, this is written as \(T(X) = AX\).
Linear transformations have the following properties:
- Additivity: For any vectors \(u\) and \(v\), a transformation \(T\) satisfies \(T(u + v) = T(u) + T(v)\).
- Homogeneity: For any vector \(v\) and scalar \(c\), \(T(cv) = cT(v)\).
Matrix Inverses
Understanding inverses is critical when solving systems of linear equations. If we have an equation \(AX = B\), the solution \(X\) can be found by multiplying both sides by \(A^{-1}\):\[ X = A^{-1}B \]Here are some important properties and facts about matrix inverses:
- Not all matrices have inverses. A matrix has an inverse if and only if it is non-singular, meaning its determinant is not zero.
- Unique Solution: If an inverse exists, the system of equations has exactly one solution.
- Reversibility: Applying the inverse transformation to the original transformation yields the original input.
Homomorphism
When we define a map \(\phi: G \longrightarrow GL_n(\mathbb{R})\) by sending each element \(\sigma_{A, B}\) to \(A\), this map is a **homomorphism** if it satisfies:\[ \phi(\sigma_{A_1, B_1} \circ \sigma_{A_2, B_2}) = \phi(\sigma_{A_1, B_1}) \cdot \phi(\sigma_{A_2, B_2}) \]This means applying the map to the composition of two functions from \(G\) results in the same matrix as multiplying the matrix representations of each function.
Homomorphisms are essential because:
- They maintain the underlying operation of the structures involved.
- They help in mapping complex structures into simpler forms, making analysis easier.
Kernel of a Map
In the context of the exercise, the kernel of the homomorphism \(\phi: G \longrightarrow GL_n(\mathbb{R})\) is given by:\[ \ker(\phi) = \{ \sigma_{A, B} \in G \ | \ \phi(\sigma_{A, B}) = I_n \} \]This means the kernel consists of all transformations where the matrix part is the identity matrix \(I_n\).
The kernel is significant because:
- It reveals the transformations that do not affect the target space's structure.
- In terms of linear algebra, it identifies solutions that map back to zero, helping to solve homogeneous equations.
- The "size" or dimension of the kernel gives information about the transformation's potential lack of injectivity.