## Linear Operators: Spectral theory |

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Page 876

Then ( y + Nie ) ( 2 ) y ( 2 ) + Ni = i ( 1 + N ) , and and

+ Nae < \ y ? + N2 . Since this inequality must hold for all real N , a contradiction ...

Then ( y + Nie ) ( 2 ) y ( 2 ) + Ni = i ( 1 + N ) , and and

**hence**11 + N = \ y + Niel .**Hence**( 1 + N ) 2 = ly + Niel2 = | ( y + Nie ) ( y + Nie ) * 1 = ( y + Nie ) ( y - Nie ) \ y2+ Nae < \ y ? + N2 . Since this inequality must hold for all real N , a contradiction ...

Page 1027

Conversely , suppose that a non - zero scalar 4 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

Conversely , suppose that a non - zero scalar 4 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**to the nullspace of T. Let u = 1 - y .Page 1227

and D_ are clearly linear subspaces of D ( T * ) , 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 xe Dr.**Hence**D is closed . Similarly , D_ is closed . Since Dtand D_ are clearly linear subspaces of D ( T * ) , 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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