commutativity theorem
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Filomat ◽  
2019 ◽  
Vol 33 (9) ◽  
pp. 2695-2704
Author(s):  
Ramesh Yousefi ◽  
Mansour Dana

A Drazin invertible operator T ? B(H) is called skew D-quasi-normal operator if T* and TTD commute or equivalently TTD is normal. In this paper, firstly we give a list of conditions on an operator T; each of which is equivalent to T being skew D-quasi-normal. Furthermore, we obtain the matrix representation of these operators. We also develop some basic properties of such operators. Secondly we extend the Kaplansky theorem and the Fuglede-Putnam commutativity theorem for normal operators to skew D-quasi-normal matrices.


2015 ◽  
Vol 61 (1) ◽  
pp. 97-100
Author(s):  
Junchao Wei ◽  
Zhiyong Fan

Abstract Let R be a ring with an identity and for each x ∈ SN(R) = {x ∈ R|x ∉ N(R)} and y ∈ R, (xy)k = xkyk for three consecutive positive integers k. It is shown in this note that R is a commutative ring, which generalizes the known theorem belonging to Ligh and Richoux.


2014 ◽  
Vol 63 (3) ◽  
pp. 355-362 ◽  
Author(s):  
B. P. Duggal ◽  
S. V. Djordjević ◽  
I. H. Jeon

Filomat ◽  
2007 ◽  
Vol 21 (2) ◽  
pp. 77-83
Author(s):  
B.P. Duggal

For Hilbert space operators A and B, let ?AB denote the generalized derivation ?AB(X) = AX - XB and let /\AB denote the elementary operator rAB(X) = AXB-X. If A is a pk-quasihyponormal operator, A ? pk - QH, and B*is an either p-hyponormal or injective dominant or injective pk - QH operator (resp., B*is an either p-hyponormal or dominant or pk - QH operator), then ?AB(X) = 0 =? SA*B*(X) = 0 (resp., rAB(X) = 0 =? rA*B*(X) = 0). .


2002 ◽  
Vol 0048 ◽  
pp. 35-42
Author(s):  
W. J. Ricker ◽  
M. Väth

1996 ◽  
Vol 30 (3-4) ◽  
pp. 302-309
Author(s):  
Jairo Z. Gonçalves ◽  
Arnaldo Mandel

1993 ◽  
Vol 24 (1) ◽  
pp. 29-36
Author(s):  
THOMAS P. KEZLAN

The purpose of this paper is to show how a previous commutativity theorem for general rings can be used to prove commutativity theorems for rings with unity, and to obtain several new results via this route, e.g., if a ring with unity satisfies either $x^k[x^n, y] = [x, y^m]x^\ell$ or $x^k[x^n,y] = [x,y^m]y^\ell (m > 1)$ and if either (A) $m$ and $n$ are relatively prime or (B) $n[x,y]=0$ implies $[x,y]=0$, then $R$ is commutative.


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