/*! 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 17 The group of all non-singular \(... [FREE SOLUTION] | 91Ó°ÊÓ

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The group of all non-singular \(n \times n\) matrices is known as the general linear group \(G L(n)\) and that with only real elements as \(G L(n, \mathbf{R}) .\) If \(\mathbf{R}^{*}\) denotes the multiplicative group of non-zero real numbers, prove that the mapping \(\Phi\) : \(G L(n, \mathbf{R}) \rightarrow \mathbf{R}^{*}\), defined by \(\Phi(\mathrm{M})=\operatorname{det} \mathrm{M}\), is a homomorphism. Show that the kernel \(\mathcal{K}\) of \(\Phi\) is a subgroup of \(G L(n, \mathbf{R})\). Determine its cosets and show that they themselves form a group.

Short Answer

Expert verified
The mapping \(\Phi(M) = \operatorname{det}(M)\) is a homomorphism. The kernel \(\mathcal{K}\) of \(\Phi\) is a subgroup of \(GL(n, \mathbf{R})\). Its cosets form a group.

Step by step solution

01

Define the Mapping

Consider the mapping \(\Phi: GL(n, \mathbf{R}) \rightarrow \mathbf{R}^*\) defined by \(\Phi(M) = \operatorname{det}(M)\). We'll show that this mapping is a homomorphism.
02

Show Homomorphism Property

To show that \(\Phi\) is a homomorphism, we need to verify that \(\Phi(AB) = \Phi(A) \Phi(B)\) for any \(A, B \in GL(n, \mathbf{R})\). Note that \(\Phi(AB) = \operatorname{det}(AB)\). Using the property of determinants, \(\operatorname{det}(AB) = \operatorname{det}(A) \times \operatorname{det}(B)\). Therefore, \(\Phi(AB) = \operatorname{det}(A) \times \operatorname{det}(B) = \Phi(A) \Phi(B)\). This confirms that \(\Phi\) is a homomorphism.
03

Define the Kernel

The kernel \(\mathcal{K}\) of \(\Phi\) is defined as the set of all matrices \(M \) in \(GL(n, \mathbf{R})\) such that \(\Phi(M) = 1\). So, \(\mathcal{K} = \{ M \in GL(n, \mathbf{R}) \mid \operatorname{det}(M) = 1 \}\), which is the set of all matrices with determinant 1.
04

Verify Subgroup Property

To verify that \(\mathcal{K}\) is a subgroup, we need to check the subgroup criteria: closure, identity, and inverse. 1. **Closure**: If \(A, B \in \mathcal{K}\), then \(\operatorname{det}(A) = 1\) and \(\operatorname{det}(B) = 1\). Therefore, \(\operatorname{det}(AB) = \operatorname{det}(A) \operatorname{det}(B) = 1 \). Thus, \(AB \in \mathcal{K}\). 2. **Identity**: The identity matrix \(I\) in \(GL(n, \mathbf{R})\) has \(\operatorname{det}(I) = 1\), so \(I \in \mathcal{K}\). 3. **Inverse**: If \(A \in \mathcal{K}\), then \(\operatorname{det}(A) = 1\). The determinant of the inverse is \(\operatorname{det}(A^{-1}) = \operatorname{det}(A)^{-1} = 1\), hence the inverse \(A^{-1} \in \mathcal{K}\). Therefore, \(\mathcal{K}\) is a subgroup.
05

Determine Cosets

The cosets of \(\mathcal{K}\) in \(GL(n, \mathbf{R})\) are of the form \(A \mathcal{K}\) for \(A \in GL(n, \mathbf{R})\). Each coset corresponds to a non-zero real number \(\Phi(A)\), since \(\Phi(A) = \operatorname{det}(A)\) and \(\Phi(K) = 1\) for all \(K \in \mathcal{K}\). Thus, the cosets can be regarded as \(\{A \mathcal{K} \mid \operatorname{det}(A) = c \}, \) with \(c \in \mathbf{R}^*\).
06

Show Cosets Form a Group

To show that the cosets form a group, consider the set of all cosets \(\{A \mathcal{K} \mid \operatorname{det}(A) = c \}\). We define multiplication of cosets: \((A \mathcal{K})(B \mathcal{K}) = AB \mathcal{K}\). For any cosets \(A \mathcal{K}\) and \(B \mathcal{K}\), \(\operatorname{det}(AB) = \operatorname{det}(A) \operatorname{det}(B) = (c \times d)\), which is again a non-zero real number. This set with coset multiplication satisfies the group axioms (closure, identity, inverses, associativity) and forms a group under this operation.

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Key Concepts

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

Nonsingular Matrices
A matrix is called 'nonsingular' or 'invertible' if it has a well-defined inverse. Mathematically, an \( n \times n \) matrix \( M \) is nonsingular if there exists another \( n \times n \) matrix \( M^{-1} \) such that \( M \times M^{-1} = I \), where \( I \) is the identity matrix. For a matrix, being nonsingular implies that it has a non-zero determinant. This property is crucial because it ensures the matrix can be used in various matrix operations, such as solving systems of linear equations. Nonsingular matrices are the backbone of many advanced topics in linear algebra and group theory.
Matrix Determinant
The determinant of a matrix is a scalar value that provides important information about the matrix. For an \( n \times n \) matrix \( M \), the determinant, denoted as \( \operatorname{det}(M) \), represents a function that maps the matrix to a real number. The determinant can be calculated using various methods, such as cofactor expansion or row reduction. Some key properties of the determinant include:
  • \( \operatorname{det}(A B) = \operatorname{det}(A) \operatorname{det}(B) \)\
  • If \( \operatorname{det}(M) = 0 \), the matrix \( M \) is singular (non-invertible).
  • \( \operatorname{det}(M^{-1}) = \operatorname{det}(M)^{-1} \) if \( M \) is nonsingular.
Determinants play a crucial role in defining matrix homomorphisms and understanding the properties of matrix groups.
Group Theory
Group theory is a branch of mathematics that studies the algebraic structures known as groups. A group is a set equipped with a binary operation satisfying four properties: closure, associativity, identity, and inverses. For example, in the context of linear algebra, the set of all nonsingular \( n \times n \) matrices forms a group under matrix multiplication. This group is known as the General Linear Group, denoted \( GL(n, \mathbb{R}) \). Understanding group theory helps in analyzing the algebraic structures that arise in various mathematical and physical contexts. Homomorphisms, which preserve the group operation, are critical tools in group theory.
Kernel
The kernel of a group homomorphism is the set of elements in the domain that map to the identity element of the codomain. For the homomorphism \( \Phi: GL(n, \mathbf{R}) \rightarrow \mathbf{R}^* \) defined by \( \Phi(M) = \operatorname{det}(M) \), the kernel \( \mathcal{K} \) consists of all matrices with determinant 1:
  • \( \mathcal{K} = \{ M \in GL(n, \mathbf{R}) \mid \operatorname{det}(M) = 1 \} \)
This set \( \mathcal{K} \) is a subgroup of \( GL(n, \mathbf{R}) \) because it satisfies the necessary conditions: closure under multiplication, existence of identity, and existence of inverses. The kernel provides insight into the structure and properties of the homomorphism.
Cosets
In group theory, cosets are a way of partitioning a group into disjoint subsets. Given a subgroup \( H \) of a group \( G \), a left coset of \( H \) in \( G \) is a subset of the form \( gH = \{ gh \mid h \in H \} \) for some \( g \in G \). For the homomorphism \( \Phi: GL(n, \mathbf{R}) \rightarrow \mathbf{R}^* \), the cosets of \( \mathcal{K} \) inside \( GL(n, \mathbf{R}) \) can be written as:
  • \( A \mathcal{K} \) where \( \operatorname{det}(A) = c \in \mathbf{R}^* \)
These cosets divide the general linear group based on the value of the determinant. The set of cosets \( \{ A \mathcal{K} \mid \operatorname{det}(A) = c \} \) forms a group under coset multiplication: \( (A \mathcal{K})(B \mathcal{K}) = AB \mathcal{K} \), illustrating how group structure is preserved in the cosets.

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Most popular questions from this chapter

Which of the following relationships between \(X\) and \(Y\) are equivalence relations? Give a proof of your conclusions in each case: (a) \(X\) and \(Y\) are integers and \(X-Y\) is odd; (b) \(X\) and \(Y\) are integers and \(X-Y\) is even; (c) \(X\) and \(Y\) are people and have the same postcode, (d) \(X\) and \(Y\) are people and have a parent in common; (e) \(X\) and \(Y\) are people and have the same mother; (f) \(X\) and \(Y\) are \(n \times n\) matrices satisfying \(Y=P X Q\), where \(P\) and \(Q\) are elements of a group \(\mathcal{G}\) of \(n \times n\) matrices.

Identify the eight symmetry operations on a square. Show that they form a group (known to crystallographers as \(4 \mathrm{~mm}\) or to chemists as \(C_{4 c}\) ) having one element of order 1, five of order 2 and two of order \(4 .\) Find its proper subgroups and the corresponding cosets.

The group of reflection-rotation symmetries of a square is known as \(\mathcal{D}_{4} ;\) let \(X\) be one of its elements. Consider a mapping \(\Phi: \mathcal{D}_{4} \rightarrow S_{4}\), the permutation group on four objects, defined by \(\Phi(X)=\) the permutation induced by \(X\) on the set \(\\{x, y, d, d\\}\), where \(x\) and \(y\) are the two principal axes and \(d\) and \(d^{\prime}\) the two principal diagonals, of the square. For example, if \(R\) is a rotation by \(\pi / 2, \Phi(R)=(12)(34)\). Show that \(\mathcal{D}_{4}\) is mapped onto a subgroup of \(S_{4}\) and, by constructing the multiplication tables for \(\mathcal{D}_{4}\) and the subgroup, prove that the mapping is a homomorphism.

The group of rotations (excluding reflections and inversions) in three dimensions that take a cube into itself is known as the group 432 (or \(O\) in the usual chemical notation). Show by each of the following methods that this group has 24 elements. (a) Identify the distinct relevant axes and count the number of qualifying rotations about each. (b) The orientation of the cube is determined if the directions of two of its body diagonals are given. Consider the number of distinct ways in which one body diagonal can be chosen to be 'vertical' and a second diagonal made to lie along a particular direction.

The following is a 'proof' that reflexivity is an unnecessary axiom for an equivalence relation. Because of symmetry \(X \sim Y\) implies \(Y \sim X .\) Then by transitivity \(X \sim Y\) and \(Y \sim X\) imply \(X \sim X .\) Thus symmetry and transitivity imply reflexivity, which therefore need not be separately required. Demonstrate the flaw in this proof using the set consisting of all real numbers plus the number \(i .\) Show by investigating the following specific cases that, whether or not reflexivity actually holds, it cannot be deduced from symmetry and transitivity alone. (a) \(X \sim Y\) if \(X+Y\) is real. (b) \(X \sim Y\) if \(X Y\) is real.

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