What has any operator got to do with dimensionality of the Hilbert space in which it is represented? Does a finite dimensional Hilbert space make sure that we have independent eigenvectors for the eigenvalues? If the Hilbert space is not finite, then also we will have infinite number of eigenvectors corresponding to different eigenvalues, isn't it?
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I am not quite sure what you are asking. Your first sentence is just far too broad a question. What exactly do you mean by "independent eigenvectors"? linearly independent? non-degenerate? something else? A particular class of operators (say Hermitian operators) or a general operator (which may or may not be diagonalizable)? – By Symmetry Jul 04 '18 at 16:57
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I am talking about operators in general. By independent eigenvectors I mean linearly independent – Cool_5275 Jul 04 '18 at 16:58
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For the n dimensional case when you have n different eigenvalues, the eigenvectors are unique. However if an eigenvalue has a multiplicity of k > 1, all you have is a k dimensional subspace and any k linearly independent vectors will serve as eigenvectors. This concept generalizes to infinite dimensional space. – herb steinberg Jul 04 '18 at 21:36