Linear Operators: Spectral theory |
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Page 876
Then ( y + Nie ) ( a ) = y ( a ) + Ni = i ( 1 + N ) , and hence 11 + N Sly + Niel .
Hence ( 1 + N ) 2 < \ y + Nie | 2 = | ( y + Nie ) ( y + Nie ) * 1 = | ( y + Nie ) ( y - Nie )
= \ y2 + Nael \ y2I + N2 . Since this inequality must hold for all real N , a
contradiction is ...
Then ( y + Nie ) ( a ) = y ( a ) + Ni = i ( 1 + N ) , and hence 11 + N Sly + Niel .
Hence ( 1 + N ) 2 < \ y + Nie | 2 = | ( y + Nie ) ( y + Nie ) * 1 = | ( y + Nie ) ( y - Nie )
= \ y2 + Nael \ y2I + N2 . Since this inequality must hold for all real N , a
contradiction is ...
Page 1027
5 shows that a is an eigenvalue and hence for some non - zero æ in H we have
Tx = ax , and hence , since T = TE , we have ( ET ) ( Ex ) = 1 Ex . Hence a belongs
to the spectrum of ET . Conversely , suppose that a non - zero scalar 1 belongs ...
5 shows that a is an eigenvalue and hence for some non - zero æ in H we have
Tx = ax , and hence , since T = TE , we have ( ET ) ( Ex ) = 1 Ex . Hence a belongs
to the spectrum of ET . Conversely , suppose that a non - zero scalar 1 belongs ...
Page 1227
Hence T * x = ix , or x € Dt . Hence Dt is closed . Similarly , D is closed . Since D
and D _ are clearly linear subspaces of D ( 7 ' * ) , it remains to show that the
spaces D ( T ) , Dr , and D are mutually orthogonal , and that their sum is D ( T * ) .
Hence T * x = ix , or x € Dt . Hence Dt is closed . Similarly , D is closed . Since D
and D _ are clearly linear subspaces of D ( 7 ' * ) , it remains to show that the
spaces D ( T ) , Dr , and D are mutually orthogonal , and that their sum is D ( T * ) .
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Contents
IX | 859 |
Eigenvalues and Eigenvectors | 903 |
Spectral Representation | 911 |
Copyright | |
15 other sections not shown
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