linear algebra - Is there a way to find $B,C$ such that $A
How to prove that every matrix with trace zero can be There are several proofs of this nice result, differing in style and applicability. Some of them work for all ground fields, some for fields of characteristic [math]0[/math], and some only for [math]\R[/math] or [math]\C[/math]. I personally prefe TRACELESS MATRICES THAT ARE NOT COMMUTATORS Traceless Matrices that are not Commutators Thesis directed by Associate Professor of Mathematics, Dr. Zachary Mesyan ABSTRACT. By a classical result, for any field F and a positive integer n, a matrix in M n(F) is a commutator if only and if it has trace zero. This is no longer true if F is replaced with an arbitrary ring R. But the only An Example of a Matrix that Cannot Be a Commutator Let I be the 2 by 2 identity matrix. Then we prove that -I cannot be a commutator of two matrices with determinant 1. That is -I is not equal to ABA^{-1}B^{-1}. Representations of Matrix Lie Algebras
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I'll assume a square matrix with real entries in my answer. 1) A matrix with trace zero has both positive and negative eigenvalues, except if the matrix is the zero matrix. This is because the trace of a matrix is equal to the sum of its eigenva Albert , Muckenhoupt : On matrices of trace zeros. On commutators of matrices over unital rings Kaufman, Michael and Pasley, Lillian, Involve: A Journal of Mathematics, 2014; Identities for the zeros of entire functions of finite rank and spectral theory Anghel, N., Rocky Mountain Journal of Mathematics, 2019; Characterization and Computation of Matrices of Maximal Trace Over Rotations Bernal, Javier and Lawrence, Jim, Journal of Geometry and How to prove that every matrix with trace zero can be
In mathematics, the special linear group SL(n, F) of degree n over a field F is the set of n × n matrices with determinant 1, with the group operations of ordinary matrix multiplication and matrix inversion.This is the normal subgroup of the general linear group given by the kernel of the determinant: (,) → ×. where we write F × for the multiplicative group of F (that is, F excluding 0).
linear algebra - Is there a way to find $B,C$ such that $A $\begingroup$ Can you explain why $\mathfrak{sl}_2(\mathbb{C})$ being semisimple implies that every matrix can be expressed as a commutator? I can see that implies it can be written as a sum of commutators, but not why it should be a single one. $\endgroup$ – Nate Mar 28 '14 at 4:02 Generalized commutators in matrix rings: Linear and Jan 19, 2015 Generalized commutators in matrix rings | Request PDF Indeed, for any n ≥ 2, there exists an n × n traceless matrix over some commutative ring S that is not a generalized commutator (respectively, is a generalized commutator but not a commutator