polynomial of M. For traceless matrices such as M 2su.N/, the coefficient a N−1 in equation (25) is equal to zero since it equals the trace of M. According to equation (25), any power N0>Nof the matrix M is identical to a linear combination of its powers Mn with 0 6n6N−1. The expansion of a matrix exponential can thus be written exp[−iM

The special linear group consists of the matrices which do not change volume, while the special linear Lie algebra is the matrices which infinitesimally do not change volume. In fact, there is an internal direct sum decomposition gl n = sl n ⊕ k of operators/matrices into traceless operators/matrices and scalars operators/matrices. The commutators, and applications case of q-traceless matrices X,Y. Some applications of the determinantal formulas are given in Sections 5 and 6. In Section 5, we How to Multiply Matrices. A Matrix is an array of numbers: A Matrix (This one has 2 Rows and 3 Columns) To multiply a matrix by a single number is easy: The kernel of this map, a matrix whose trace is zero, is often said to be traceless or tracefree, and these matrices form the simple Lie algebra sl n, which is the Lie algebra of the special linear group of matrices with determinant 1.

This gives, in particular, a strong negative solution to the problem whether n × n traceless matrices are necessarily commutators over a commutative base ring, for any n ≥ 2. Keywords: matrix rings , commutators , generalized commutators , traceless matrices , elementary divisor rings

The inner product in this space is defined as: $(A,B)=Trace(A^\dagger B)$,where A,B are 2x2 traceless Hermitian matrices. In order to prove that the above vector space is a Hilbert space, we can consider a Cauchy sequence and show that it converges into the space. Mar 31, 2015 · I read the following as a model solution to a question but I don't understand it - " there is no possible finite dimensional representation of the operators x and p that can reproduce the commutator [x,p] = I(hbar)(identity matrix) since the LHS has zero trace and the RHS has finite trace. My

$\begingroup$ I see, but wouldn't this show that traceless matrices are sums of commutators? I want to know wether any traceless matrix is a commutator. I'll edit my question to make it clearer. $\endgroup$ – Olivier Bégassat Mar 27 '12 at 21:57

This gives, in particular, a strong negative solution to the problem whether n × n traceless matrices are necessarily commutators over a commutative base ring, for any n ≥ 2. Keywords: matrix rings , commutators , generalized commutators , traceless matrices , elementary divisor rings