نتایج جستجو برای: commutator
تعداد نتایج: 1974 فیلتر نتایج به سال:
Theorem ([11, Theorem 2, page 282]). Let R be a prime ring, L a noncommutative Lie ideal of R and d 6= 0 a derivation of R. If [d(x), x] ∈ Z(R), for all x ∈ L, then either R is commutative, or char(R) = 2 and R satisfies s4, the standard identity in 4 variables. Here we will examine what happens in case [d(x), x]n ∈ Z(R), for any x ∈ L, a noncommutative Lie ideal of R and n ≥ 1 a fixed integer....
We prove that for any odd integer N and any integer n > 0, the Nth power of a product of n commutators in a nonabelian free group of countable infinite rank can be expressed as a product of squares of 2n+1 elements and, for all such odd N and integers n, there are commutators for which the number 2n+1 of squares is the minimum number such that the Nth power of its product can be written as a pr...
If 3C is a separable (complex) Hubert space, and A is a (bounded, linear) operator on 3C, then A is a commutator if there exist operators B and C on 3C such that 4 = BCCB. I t was shown by Wintner [8] and also by Wielandt [7] that no nonzero scalar multiple of the identity operator I on 3C is a commutator, and this was improved by Halmos [5] who showed that no operator of the form X / + C is a ...
The operators on `∞ which are commutators are those not of the form λI + S with λ 6= 0 and S strictly singular.
We show that elements of unital C-algebras without tracial states are finite sums of commutators. Moreover, the number of commutators involved is bounded, depending only on the given C-algebra.
The operators on Lp = Lp[0, 1], 1 ≤ p < ∞, which are not commutators are those of the form λI + S where λ , 0 and S belongs to the largest ideal in L(Lp). The proof involves new structural results for operators on Lp which are of independent interest.
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