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This article first summarize the corresponding results from the matrix case before discussing the spectral properties of compact operators. The reader will see that most statements transfer verbatim from the matricial case. The spectral theory of compact operators was first developed by F. Riesz . SPECTRAL THEORY OF MATRICES The classification result for square matrices is the Jordan canonical form, which states the following: Theorem Let ''A'' be an ''n'' × ''n'' complex matrix, i.e. ''A'' a linear operator acting on '''C'''''n''. If ''λ''1...''λk'' are the distinct eigenvalues of ''A'', then '''C'''''n'' can be decomposed into the invariant subspaces of ''A'' : The subspace ''Yi'' = ''Ker''(''λi'' - ''A'')''m'' where ''Ker''(''λi'' - ''A'')''m'' = ''Ker''(''λi'' - ''A'')''m''+1. Furthermore, the poles of the resolvent function ''ζ'' → (''ζ'' - ''A'')-1 coincide with the set of eigenvalues of ''A''. COMPACT OPERATORS Statement Let ''X'' be a Banach space, ''C'' be a compact operator acting on ''X'', and ''σ''(''C'') be the Spectrum of ''C''. The spectral properties of ''C'' are: Theorem i) Every nonzero ''λ'' ∈ ''σ''(''C'') is an eigenvalue of ''C''. ii) For all nonzero ''λ'' ∈ ''σ''(''C''), there exist ''m'' such that ''Ker''(''λi'' - ''A'')''m'' = ''Ker''(''λi'' - ''A'')''m''+1. iii) The eigenvalues can only accumulate at 0. If the dimension of ''X'' is not finite, then ''σ''(''C'') must accumulate at 0. iv) ''σ''(''C'') is countable. v) Every nonzero ''λ'' ∈ ''σ''(''C'') is a pole of the resolvent function ''ζ'' → (''ζ'' - ''C'')-1. Proof The theorem claims several properties of the operator ''λ'' - ''C'' where ''λ'' ≠ 0. Without loss of generality, it can be assumed that ''λ'' = 1. Therefore we consider ''I'' - ''C'', ''I'' being the identity operator. The proof will require two lemmas. The first is called Riesz's Lemma : | ||
| But ''d''(''x<sub>n</sub>'', ''Ker''(''I'' - ''C'')) Must Be Bounded Suppose This Is Not The Case Pass Now To The Quotient Map Of (''I'' - ''C''), Still Denoted By (''I'' - ''C''), On ''X''/''Ker''(''I'' - ''C'') The Quotient Norm On ''X''/''Ker''(''I'' - ''C'') Is Still Denoted By ·, And {''x<sub>n</sub>''} Are Now Viewed As Representatives Of Their Equivalence Classes In The Quotient Space Take A Subsequence {''x<sub>n K</sub>''} Such That ''x<sub>n K</sub>'' > K And Define A Sequence Of Unit Vectors By ''z<sub>n K</sub>'' | ''x<sub>n k</sub>''/''x<sub>n k</sub>'' Again we would have (''I'' - ''C'')''z<sub>n k</sub>'' &rarr (''I'' - ''C'')''z'' for some ''z'' Since (''I'' - ''C'')''z<sub>n k</sub>'' = (''I'' - ''C'')''x<sub>n k</sub>''/ ''x<sub>n k</sub>'' &rarr 0, we have (''I'' - ''C'')''z'' = 0 ie ''z'' &isin ''Ker''(''I'' - ''C'') Since we passed to the quotient map, ''z'' = 0 This is impossible because ''z'' is the norm limit of a sequence of unit vectors Thus the lemma is proven |
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| :<math>\ C Y N - C Y M \ | \ (C-I) y_n + y_n - (C-I) y_m - y_m \</math> |
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| :<math>\ C Y N - C Y M \ | \ (C-I) y_n + y_n - (C-I) y_m - y_m \</math> |
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| Which Implies ''Cy<sub>n</sub> - Cy<sub>m</sub>'' > ½ This Is A Contradiction, And So The Sequence { ''Y<sub>n</sub>'' | ''Ker''(''λ<sub>i</sub>'' - ''A'')<sup>''n''</sup>} must terminate at some finite ''m'' |
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| '''iii''') Suppose Eigenvalues Of ''C'' Do Not Accumulate At 0 We Can Therefore Assume That There Exist A Sequence Of Distinct Eigenvalues {''λ<sub>n</sub>''}, With Corresponding Eigenvectors {''x<sub>n</sub>''}, Such That ''λ<sub>n</sub>'' > ''ε'' For All ''n'' Define ''Y<sub>n</sub>'' | ''span''{''x''<sub>1</sub>''x<sub>n</sub>''} The sequence {''Y<sub>n</sub>''} is a strictly increasing sequence Choose unit vectors such that ''y<sub>n</sub>'' &isin ''Y''<sup>''n''</sup> and ''d''(''y<sub>n</sub>'', ''Y''<sup>''n'' - 1</sup>) = 1 Then for ''n'' < ''m'' |
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| \ C Y N - C Y M \ | \ (C- \lambda_n) y_n + \lambda_n y_n - (C- \lambda_m) y_m - \lambda_m y_m \ |
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| :<math>\cup N \{ \lambda > Rac{1}{n} \} | \cup_n S_n </math> |
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