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I'm reading the Algebra book by Knapp and he mentions in passing that an orthonormal set in an infinite dimension vector space is "never large enough" to be a vector-space basis (i.e. that every vector can be written as a finite sum of vectors from the basis; such bases do exist for infinite dimension vector spaces by virtue of the axiom of choice, but usually one works with orthonormal sets for which infinite sums yield all the elements of the vector space).

So my question is - how can we prove this claim? (That an orthonormal set in an infinite dimension vector space is not a vector space basis).


Edit (by Jonas Meyer): Knapp says in a footnote on page 92 of Basic algebra:

In the infinite-dimensional theory the term "orthonormal basis" is used for an orthonormal set that spans $V$ when limits of finite sums are allowed, in addition to finite sums themselves; when $V$ is infinite-dimensional, an orthonormal basis is never large enough to be a vector-space basis.

Thus, without explicitly using the word, Knapp is referring only to complete infinite-dimensional inner product spaces.

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I may be missing something here, but a Fourier basis is certainly looks like an orthonormal set to me, therefore... where is the problem? Or is it just because you are excluding infinite sums? –  Raskolnikov Dec 9 '10 at 12:02
    
Note that to define the notion of orthonormal, you need some inner product on the space. Therefore let us restrict to Hilbert spaces. The problem here is that the linear span of the orthonormal basis is only dense in the Hilbert space. –  user1119 Dec 9 '10 at 12:21
    
I think the question is about finite sums, i.e. about Hamel bases. –  Carl Mummert Dec 9 '10 at 12:22
    
Indeed, Fourier basis is NOT a "vector space basis" since we need an infinite sum to represent all the elements in the space. –  Gadi A Dec 9 '10 at 14:21
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Given the clarification provided by the answers, it would be interesting to know exactly what Knapp says in his book. Could someone provide a direct quote? –  Pete L. Clark Dec 9 '10 at 18:31

3 Answers 3

up vote 15 down vote accepted

Edit. The answer assumes that one works in a Hilbert space.

Let $\mathcal A$ denote the set of orthonormal vectors. Then for any orthonormal sequence $(e_n)\subset\mathcal A $, the element $$u=\sum_{n=1}^{\infty}\frac{1}{n}e_n$$ cannot be represented as a finite sum of vectors from $\mathcal A$.

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Why is this necessarily an element of your vector space? Vector spaces are typically only closed under finite sums. Then again, I'm very used to working with finite dimensional vectors spaces, so I could just be plain wrong. –  Jason DeVito Dec 9 '10 at 12:03
    
I agree with Jason, take for instance the vector space of all polynomials. –  Raskolnikov Dec 9 '10 at 12:05
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I think the problem may be the following: that the OP speaks of orthonormal implies he is working with an inner product space (in particular, a metric space). Perhaps the statement in the book should be properly interpreted as a statement about complete inner product spaces? The set of finite length vectors in $\ell^2$ is not complete. –  Willie Wong Dec 9 '10 at 12:21
    
Thanks to everyone for the comments. I implicitly assumed that the space is complete. –  Andrey Rekalo Dec 9 '10 at 12:35
    
How do you know that the partial sums are Cauchy? We don't know with which norm we are working with(i.e. not necessarily the standard metric). –  TKM Jan 15 at 21:35

Take any infinite dimensional inner product space $V$ and any orthonormal sequence $(w_n : n \in \mathbb{N})$. Let $W$ be the subspace generated by this sequence. Then $W$ is certainly an infinite dimensional vector space (because it has an infinite independent subset). Also $W$ has an orthonormal basis, because the inner product on $W$ is inherited from $V$ and thus $(w_n)$ is still an orthonormal sequence in $W$. This means that the theorem you have suggested, "an orthonormal set in an infinite dimension vector space is not a vector space basis", is not true.

What I believe might be true is that no infinite dimensional complete inner product space has a orthonormal basis. This is the question that Andrey Rekalo addressed in another answer.

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Indeed, if completeness is required, I agree with the statement. –  Raskolnikov Dec 9 '10 at 15:15

As already mentioned in the other answers and comments, this is true if the space is assumed to be complete (see Andrey Rekalo's answer), but not necessarily otherwise (see Carl Mummert's answer). In the complete case, more can be said. If the Hilbert space dimension (cardinality of a maximal orthonormal set) is infinite, then the linear dimension (cardinality of a maximal linearly independent set) is at least $\mathfrak{c}=2^{\aleph_0}=|\mathbb{R}|$. Of course, this doesn't directly tell you anything if the Hilbert space dimension is $\mathfrak{c}$ or greater, but in the case of separable Hilbert spaces like $\ell^2$ (as defined here), it tells you that not only does an orthonormal set not span, but no subset of cardinality less than $\mathfrak{c}$ can span. One way to see this is to consider the linearly independent set $\{(1,t,t^2,t^3,\ldots):-1\lt t\lt 1\}\subset\ell^2$. Since $\ell^2$ imbeds into every infinite dimensional Hilbert space as the closed linear span of any countably infinite orthonormal set, this also demonstrates the general fact. This and other proofs can be found in the solutions to Problem 7 of Halmos's Hilbert Space Problem Book, which I highly recommend. This statement also extends to Banach spaces.

Strictly speaking, what I have said so far hasn't answered your question in the case of Hilbert space dimension greater than or equal to $\mathfrak{c}$. However, you can get an overkill solution to your question by taking a countably infinite subset $A$ of an orthonormal subset of an arbitrary Hilbert space $H$. If $M_0$ is the linear span of $A$ and $M=\overline{M_0}$ is the closed linear span of $A$, then $H=M\oplus M^\perp$, while the linear span of your orthonormal set is contained in $M_0\oplus M^\perp$, which is properly contained in $H$ because $M$ has higher dimension than $M_0$. ($M^\perp$ denotes the set of vectors orthogonal to every element of $M$, and $\oplus$ is being used here to denote internal direct sums.) All that was really needed here is that the linear dimension of $M$ is not countable, and this also follows from Baire's theorem.

Andrey Rekalo gives a better, nonoverkill answer for the complete case.

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