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In the chapter 6.4 on normal and self-adjoint operators, there is an example of an infinite dimensional inner product space H that has a normal operator but that has no eigenvectors.

The space is the set of functions f_n(t) = e^(int) , t in [0,2pi] with inner product = 1/2pi * integral_0_2pi(e^(-int) e^(imt))dt

The operator T(f_n) = f_(n+1)

My question is what is the basis for the validity of the following statement.

any vector f in the inner product space can be represented as f = sum_i=n_to_i=m(a_i * f_i), a_m <> 0

They use this fact to prove that the operator as no eigenvectors. But it seems to imply that any member of an infinite dimensional inner product space can be represented with a finite number of basis elements, since n and m are finite.

Are there no elements of an infinite dimensional inner product space that require an infinite number of terms to represent?

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2 Answers 2

up vote 4 down vote accepted

The space being defined is the span of the functions fn, and in the definition of the span we only allow finite sums of the basis vectors.

Edit: I should also mention that the notion of infinite sum in an inner product space doesn't make sense unless the space is also complete with respect to the induced norm, i.e. is a Hilbert space. In the nice situation a Hilbert space has an "orthonormal basis," which is not a basis in the linear algebra sense but in the sense that the span of the basis is dense.

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Qiaochu Yuan hit it on the head. To quote Hoffman & Kunze (Linear Algebra):

Infinite bases have nothing to do with 'infinite linear combinations.' The reader who feels an irresistible urge to inject power series into this example [concerning vector spaces of polynomials] should study the example carefully again. If that does not effect a cure, he should consider restricting his attention to finite-dimensional spaces from now on. [!]

That exactly describes your situation: you have a vector space of polynomials in e^(it), and you're trying to insert power series. If you go back to the definition of "vector space," you should verify that the definitions and axioms allow you to form arbitrary finite linear combinations of vectors, but not "infinite" ones. Note that in certain vector spaces, wisely constructed infinite sums are okay. Those include Banach spaces and Hilbert spaces. However, your vector space is not a Banach space or a Hilbert space, so there's no concept of "infinite sum." (You would probably learn about Hilbert spaces and Banach spaces while studying real analysis or functional analysis.)

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Thanks to both comments. This clarifies things. –  Jeff Oct 18 '09 at 2:45

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