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TENSOR PRODUCT OF CONTINUOUS OPERATOR
Author(s) -
M. Abdul–Munem Maysaa
Publication year - 2007
Publication title -
journal of al-nahrain university-science
Language(s) - English
Resource type - Journals
eISSN - 2519-0881
pISSN - 1814-5922
DOI - 10.22401/jnus.10.2.24
Subject(s) - tensor product , tensor product of hilbert spaces , tensor product of modules , tensor contraction , tensor product of algebras , cartesian tensor , tensor (intrinsic definition) , mathematics , product (mathematics) , invariant (physics) , tensor density , operator (biology) , pure mathematics , tensor field , exact solutions in general relativity , mathematical analysis , mathematical physics , geometry , biochemistry , chemistry , repressor , gene , transcription factor
In this paper we prove that some properties of tensor product and we show that if 2 1, A A are θadjoint then 2 1 A I I A is normal .Also we prove that a continuous operator is invariant under tensor product. Introduction Let H be an infinite dimensional separable complex Hilbert space with inner product , and let H B be the algebra of all bounded linear operators on H , given H B A A 2 1, , the tensor product 2 1 A A on the Hilbert space H H has been considered variously by many of authors (see [2],[3],[5],[6],[7]) . When 2 1 A A is defined as follows 2 1 A A 2 2 1 1 y x , y x 2 1 2 2 1 1 y , y A x , x A The operation of taking tensor product 2 1 A A preserves many properties of H B A and A 2 1 but by no means all of them , Thus, whereas the binormal property is invariant under tensor product ,the *paranormal property is not [9] .a gain ,whereas 2 1 A A is posinormal if and only if 1 A and 2 A H B are [9] and is similarly for Uoperator ,pseudo normal ,subnormal and normaloid operators [3],[9],[10] .it was shown in [9]that paranormal is not invariant under tensor product . In this section we prove some properties of tensor product . Proposition If 0 B A and 0 D C then 0 D B C A Proof: Sinc e 0 B A then x , Bx x , Ax H x And Since 0 D C then 1 1 1 1 x , Dx x , Cx H x1 1 1 1 1 x , Dx x , Bx x , Cx x , Ax C A 1 1 , x x x x D B 1 1 , x x x x D B C A 1 1 x x , x x 0 1 x x H H then 0 D B C A . Proposition If 0 B C A then B B C B A C 2 2 proof Since 0 B C A then 0 x , Bx x , Cx x , Ax H x And 0 x , x B x , x A 2 2 0 , , , , , , 2 2 x Bx x x B x Cx x Bx x x A x Cx B A C 2 x x x , x x x B B C 2 0 , x x x x x x B B C B A C 2 2 x x x , x x x 0 then B B C B A C 2 2 Proposition If 0 B A then

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