## Linear Operators, Part 2 |

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Page 1092

Let T be a compact operator , and 1 , ( T ) an enumeration of its eigenvalues , repeated according to multiplicity , and in decreasing order of absolute values . ( If there are only a

Let T be a compact operator , and 1 , ( T ) an enumeration of its eigenvalues , repeated according to multiplicity , and in decreasing order of absolute values . ( If there are only a

**finite**number N of non - zero eigenvalues , we write ...Page 1460

Then , if r is

Then , if r is

**finite**below a , and the leading coefficient of t + t , never vanishes , t + ty is**finite**below 2 . Proof . It is clear that we may suppose without loss of generality that 1 = 0. By Corollary 24 ( b ) , Corollary XII.4.13 ...Page 1913

( See also Decomposition ) definition , III.4.3 ( 126 )

( See also Decomposition ) definition , III.4.3 ( 126 )

**finite**, III.4.3 ( 126 ) Lebesgue extension of , III.5.18 ( 143 ) as a metric space , III.7.1 ( 158 ) , III.9.6 ( 169 ) positive , III.4.3 ( 126 ) product , of**finite**number of ...### What people are saying - Write a review

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### Contents

BAlgebras | 859 |

Bounded Normal Operators in Hilbert Space | 887 |

Miscellaneous Applications | 937 |

Copyright | |

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additive Akad algebra Amer analytic assume Banach spaces basis belongs Borel boundary conditions boundary values bounded called clear closed closure coefficients compact complex Consequently constant contains continuous converges Corollary corresponding defined Definition denote dense determined domain eigenvalues element equal equation essential spectrum evident Exercise exists extension finite follows formal differential operator formula function given Hence Hilbert space identity independent indices inequality integral interval Lemma limit linear mapping Math matrix measure multiplicity Nauk neighborhood norm obtained partial positive preceding present problem projection proof properties prove range regular remark representation respectively restriction result Russian satisfies seen sequence singular solution spectral square-integrable statement subset subspace sufficiently Suppose symmetric Theorem theory topology transform unique vanishes vector zero