The study of operators satisfying σja（T ） = σa（T ） is of significant interest. Does σja（T ） = σa（T ） for n-perinormal operator T ∈ B（H）？ This question was raised by Mecheri and Braha [Oper. Matrices 6 ...The study of operators satisfying σja（T ） = σa（T ） is of significant interest. Does σja（T ） = σa（T ） for n-perinormal operator T ∈ B（H）？ This question was raised by Mecheri and Braha [Oper. Matrices 6 （2012）, 725-734]. In the note we construct a counterexample to this question and obtain the following result： if T is a n-perinormal operator in B（H）, then σja（T ）／{0} = σa（T ）／{0}. We also consider tensor product of n-perinormal operators.展开更多
基金supported by NNSF(1122618511201126)the Basic Science and Technological Frontier Project of Henan Province(132300410261)
文摘The study of operators satisfying σja（T ） = σa（T ） is of significant interest. Does σja（T ） = σa（T ） for n-perinormal operator T ∈ B（H）？ This question was raised by Mecheri and Braha [Oper. Matrices 6 （2012）, 725-734]. In the note we construct a counterexample to this question and obtain the following result： if T is a n-perinormal operator in B（H）, then σja（T ）／{0} = σa（T ）／{0}. We also consider tensor product of n-perinormal operators.
