/*! 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 2 Show that the center of \(\mathr... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 2

Show that the center of \(\mathrm{GL}_{n}(F)\) is the group of scalar non-zero matrices.

Short Answer

Expert verified
We have shown that scalar non-zero matrices in the form of \( A = cI_n \), where \( c \neq 0 \) and \( I_n \) is the identity matrix of order \( n \), are in the center of \(\mathrm{GL}_{n}(F)\) because they commute with every element in the group. Moreover, we have shown that no other type of matrices can be in the center of \(\mathrm{GL}_{n}(F)\) since they do not commute with every element. Therefore, we can conclude that the center of \(\mathrm{GL}_{n}(F)\) is the group of scalar non-zero matrices.

Step by step solution

01

1. Definitions

We will begin by defining some terms: - The General Linear group, denoted as \(\mathrm{GL}_{n}(F)\), is the group of all invertible \(n\times n\) matrices with entries from a field \(F\). - The center of a group, denoted as \(Z(G)\), is the set of all elements in a group that commute with every other element in the group. It is defined as \(Z(G) = \{z \in G \mid zg = gz \,\, \forall g \in G\}\). - A matrix is called a scalar matrix if it is a diagonal matrix with all non-diagonal entries equal to zero, and all diagonal entries equal to a constant. Now we will proceed with the proof.
02

2. Scalar non-zero matrices are in the center

Let's take a scalar non-zero matrix \( A = cI_n \), where \( c \neq 0 \), and \( I_n \) is the identity matrix of order \( n \). For any invertible matrix \( B \in \mathrm{GL}_{n}(F) \), we need to show that \( AB = BA \). We have: \(AB = (cI_n)B = c(BI_n) = cB\) and, \(BA = B(cI_n) = (BI_n)c = Bc\) Here, both products are equal: \(AB = BA = cB\), which means that the scalar non-zero matrix \(A\) commutes with every \(B \in \mathrm{GL}_{n}(F)\). Hence, scalar non-zero matrices are in the center of \(\mathrm{GL}_{n}(F)\).
03

3. Other types of matrices are not in the center

Now, let's prove that no other type of matrices except scalar non-zero matrices can be in the center of \(\mathrm{GL}_{n}(F)\). Suppose there is such a matrix \(A\) in the center of \(\mathrm{GL}_{n}(F)\), which is not a scalar matrix. Then, there exists at least one off-diagonal entry that is non-zero. Consider a matrix \(B\) such that it has non-zero entries in the same row and column as the non-zero entry of matrix \(A\). For example, if \(A\) has a non-zero entry in the second row and the first column, then consider a matrix \(B\) with non-zero entries in the second row and the first column. Now compute the products \(AB\) and \(BA\). Since the non-zero off-diagonal entries of \(A\) and \(B\) are in the same row and column, at least one entry will differ in the resulting matrix products. Therefore, \(AB \neq BA\), and \(A\) does not commute with \(B\). We can thus conclude that no other type of matrices except scalar non-zero matrices can be in the center of \(\mathrm{GL}_{n}(F)\).
04

4. Conclusion

In this exercise, we have shown that scalar non-zero matrices are in the center of \(\mathrm{GL}_{n}(F)\) and no other type of matrices can be in the center. Therefore, the center of \(\mathrm{GL}_{n}(F)\) is indeed the group of scalar non-zero matrices.

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 91影视!

Key Concepts

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

General Linear Group
The General Linear group, often denoted by \( \mathrm{GL}_{n}(F) \), plays a fundamental role in advanced algebra and its applications. This group consists of all the invertible \( n \times n \) matrices where the entries are from a set known as a field \( F \). A field is a system consisting of elements where addition, subtraction, multiplication, and division (except division by zero) operations can be carried out.

Understanding the General Linear group helps grasp the concept of linear transformations and automorphisms in vector spaces. It's crucial to recognize that a matrix is invertible if it has a non-zero determinant, which means there's a unique matrix that, when multiplied with the original matrix, gives the identity matrix. The identity matrix \( I_n \) is the matrix with all diagonal entries as one, and the rest are zero.

The General Linear group also contains subgroups with specific properties, and understanding these subgroups, such as the center of \( \mathrm{GL}_{n}(F) \), gives insights into the symmetry and structure of linear transformations.
Scalar Matrix
Within the realm of matrices, a scalar matrix holds a special place. A matrix is called a scalar matrix if and only if it is a diagonal matrix where all diagonal entries are equal, and all other entries are zero. In other words, a scalar matrix looks like \( cI_n \), where \( c \), a non-zero scalar from the field \( F \), multiplies the identity matrix \( I_n \).

The characteristic that distinguishes scalar matrices is often described by their simplicity and their effect on other matrices when used in multiplication. They scale or stretch the other matrix by the scalar value, keeping the direction unchanged. This property of scalar matrices makes them commutative with any invertible matrix in \( \mathrm{GL}_{n}(F) \), as they essentially 'respect' the structure of the other matrix, simply applying a uniform scaling across all rows or columns.
Commutative Property in Groups
The commutative property is a well-known concept, typically introduced in elementary arithmetic, stating that the order of numbers does not affect the result of the operation, such as in addition or multiplication. When it comes to more advanced mathematical structures like groups, this property is not to be taken for granted.

In group theory, the commutative property isn't always valid. However, there is a special subset within a group where the commutative property holds for every element outside of that subset. This subset is known as the center of the group. Formally, the center \( Z(G) \) of a group \( G \) consists of all elements that commute with all other elements in \( G \), meaning for any element \( z \) in the center, and any element \( g \) in \( G \) the equation \( zg = gz \) holds true.

The importance of the center lies in its ability to harness the commutative property and facilitate understanding the internal structure and symmetry of the group. When we identify the center of \( \mathrm{GL}_{n}(F) \), for instance, we gain valuable insight into how these matrices interact with each other and the invariant transformations they represent.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(L\) be hermitian, and let \(v, v^{\prime}\) be eigenvectors with distinct eigenvalues for \(L\). Show that \(v, v^{\prime}\) are orthogonal.

Let \(A\) be the subgroup of \(G\) consisting of all matrices $$s(a)=\left(\begin{array}{cc} a & 0 \\ 0 & a^{-1} \end{array}\right) \quad \text { with } a>0 .$$ (a) Show that the map \(a \mapsto s(a)\) is a homomorphism of \(\mathbf{R}^{*}\) into \(G\). Since this homomorphism is obviously injective, this homomorphism gives an imbedding of \(\mathbf{R}^{+}\) into \(G\). (b) Let \(U\) be the subgroup of \(G\) consisting of all clements $$u(x)=\left(\begin{array}{ll} 1 & x \\ 0 & 1 \end{array}\right) \quad \text { with } x \in \mathbf{R}$$ Thus \(u \mapsto u(x)\) gives an imbedding of \(\mathbf{R}\) into \(G\). Then \(U A\) is a subset of \(G\). Show that \(U A\) is a subgroup. How does it differ from the Borel subgroup of \(G\) ? Show that \(U\) is normal in \(U A\). (c) Show that the map $$U A \rightarrow \mathbf{H}$$ given by $$\beta \mapsto \beta(i)$$ gives a bijection of \(U A\) onto \(\mathrm{H}\). (d) Show that every element of \(\mathrm{SL}_{2}(\mathbf{R})\) admits a unique expression as a product $$u(x) s(a) r(\theta),$$ so in particular, \(G=U A K\).

Given a real number 0 , let $$r(\theta)=\left(\begin{array}{ll} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right)$$ (a) Show that \(\theta \mapsto r(\theta)\) is a homomorphism of \(R\) into \(G\). We denote by \(K\) the set of all such matrices \(r(\theta)\). So \(K\) is a subgroup of \(G\). (b) Show that if \(x=r(\theta)\) then \(x(i)=i\), where \(i=\sqrt{-1}\). (c) Show that if \(\alpha \in G\), and \(\alpha(i)=i\) then there is some \(\theta\) such that \(\alpha=r(\theta)\). In the terminology of the operation of a group, we note that \(K\) is the isotropy group of \(i\) in \(G\).

Let \(F\) be the quotient field of the principal ring \(R\). Let \(B_{n}(F)\) be the subset of \(\mathrm{GL}_{m}(F)\) consisting of the upper triangular matrices (arbitrary on the diagonal, but non-zero determinant). Show by induction that $$\mathrm{GL}_{-}(F)=\mathrm{SL}_{n}(R) B_{n}(F) .$$ | Hint: First do the case \(n=2\) using Exercise 8. Next let \(n>2\). Let \(X=\left(x_{i j}\right)\) be an unknown matrix in \(\mathrm{SL}_{n}(R)\), and let \(X_{n}\) be its bottom row, Let \(A^{1} \ldots, A^{n}\) be the columns of a given matrix \(A \in \mathrm{GL}_{n}(F)\). We want to solve for \(X_{n} \cdot A^{i}=0\) for \(i=1, \ldots, n-1\), so that \(X A\) has its last row equal to 0 except for the lower right comer, Consider the \(R\) -module consisting of \(R\) -vectors \(X_{n}\) satisfying these orthogonality relations. It has a non-zero element, and by unique factorization it has an element \(X_{n}\) whose components are relatively prime. Use Exercise 8. For some \(A^{\prime} \in \mathrm{GL}_{n-1}(F), X A\) is a matrix with 0 in the bottom row except for the lower right hand corner, which is a unit in \(R\). Use induction again to get a matrix \(Y \in \mathrm{SL}_{n-1}(R)\) such that \(Y A^{\prime}\) is upper triangular. Conclude.

Let \(\zeta\) be a primitive \(n\) -th root of unity where \(n\) is an odd integer. Let \(G\) be the subgroup of all \(2 \times 2\) matrices generated by the matrices $$w=\left(\begin{array}{rr} 0 & -1 \\ 1 & 0 \end{array}\right) \quad \text { and } \quad z=\left(\begin{array}{ll} 6 & 0 \\ 0 & \zeta^{-1} \end{array}\right)$$ Show that \(G\) has order \(4 n\) What is wzw \(^{-1}\) ?
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation

Applied Mathematics

Read Explanation

Calculus

Read Explanation

Logic and Functions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.