Linear Operators, Part 2 |
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Page 876
Then ( y + Nie ) ( 2 ) = y ( 2 ) + Ni = i ( 1 + N ) , and hence 11 + NI Sly + Niel .
Hence ( 1 + N ) 2 5 \ y + Niel2 = | ( y + Nie ) ( y + Nie ) * 1 = | ( y + Nie ) ( y - Nie ) =
\ y2 + Nael = \ y2 ! + N2 . Since this inequality must hold for all real N , a
contradiction ...
Then ( y + Nie ) ( 2 ) = y ( 2 ) + Ni = i ( 1 + N ) , and hence 11 + NI Sly + Niel .
Hence ( 1 + N ) 2 5 \ y + Niel2 = | ( y + Nie ) ( y + Nie ) * 1 = | ( y + Nie ) ( y - Nie ) =
\ y2 + Nael = \ y2 ! + N2 . Since this inequality must hold for all real N , a
contradiction ...
Page 1027
Hence a belongs to the spectrum of ET . Conversely , suppose that a non - zero
scalar , belongs to the spectrum of ET . Then , for some non - zero X in EH , we
have ETx = 2x . Then Tx = 2x + y , where y belongs to the subspace ( I - E ) H ,
and ...
Hence a belongs to the spectrum of ET . Conversely , suppose that a non - zero
scalar , belongs to the spectrum of ET . Then , for some non - zero X in EH , we
have ETx = 2x . Then Tx = 2x + y , where y belongs to the subspace ( I - E ) H ,
and ...
Page 1227
Hence T * x = ix , or x e Dr . Hence D . is closed . Similarly , D _ is closed . Since
Dt and D _ are clearly linear subspaces of D ( T * ) , it remains to show that the
spaces D ( T ) , Du , and D are mutually orthogonal , and that their sum is D ( T * )
.
Hence T * x = ix , or x e Dr . Hence D . is closed . Similarly , D _ is closed . Since
Dt and D _ are clearly linear subspaces of D ( T * ) , it remains to show that the
spaces D ( T ) , Du , and D are mutually orthogonal , and that their sum is D ( T * )
.
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Contents
BAlgebras | 859 |
Commutative BAlgebras | 868 |
Commutative BAlgebras | 874 |
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