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अंग्रेजी-हिंदी > invariant subspace उदाहरण वाक्य

invariant subspace उदाहरण वाक्य

उदाहरण वाक्य
41.Any subspace spanned by eigenvectors of " T " is an invariant subspace of " T ", and the restriction of " T " to such a subspace is diagonalizable.

42.I am still not convinced that the category is coherent; for instance, the Perron Frobenius theorem is not a theorem on invariant subspaces except in the trivial sense that any eigenvector determines an invariant subspace.

43.I am still not convinced that the category is coherent; for instance, the Perron Frobenius theorem is not a theorem on invariant subspaces except in the trivial sense that any eigenvector determines an invariant subspace.

44.The invariant subspace problem concerns the case where " V " is a separable Hilbert space over the complex numbers, of dimension > 1, and " T " is a bounded operator.

45.In mathematics, a linear operator f : V \ to V is called "'locally finite "'if the space V is the union of a family of finite-dimensional f-invariant subspaces.

46.When you say " it is unknown whether all linear operators on separable complex Hilbert spaces have ( non-trivial ) invariant subspaces ", I agree with you fully, but I'm not sure of the relevance.

47.In the field of mathematics known as functional analysis, the "'invariant subspace problem "'is a partially unresolved problem asking whether every bounded operator on a complex Banach space sends some non-trivial compact square.

48.This was resolved affirmatively, for a more general class of polynomially compact operators, by Allen R . Bernstein and Abraham Robinson in 1966 ( see Non-standard analysis # Invariant subspace problem for a summary of the proof ).

49.The Bombieri inequality implies that the product of two polynomials cannot be arbitrarily small, and this lower-bound is fundamental in applications like polynomial factorization ( or in Enflo's construction of an operator without an invariant subspace ).

50.If " G " were not compact, but were abelien, then diagonalisation is not achieved, but we get a unique " continuous " decomposition of " H " into 1-dimensional invariant subspaces.

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