A basis for an infinite-dimensional vector space, by definition, must have an infinite number of vectors in it. Without more information, I'd have to say that the "result" is wrong, and your intuition correct.
I was working on a problem earlier today and I didn't know the following result:
Let S be a subset of an infinite-dimensional vector space V. Then S is a basis for V if and only if for each nonzero vector v in V, there exists unique vectors u1,u2,...,un in S and unique nonzero scalars c1,c2,...cn, such that v = (c1)u1 + (c2)u2 + ... + (cn)un.
I don't "see" how this can be true? For example, lets say I take the vector space of infinite-tuples so x = (x1, x2, ...). How is it that I can write this as a linear combination of FINITE number of elements of S (a basis of this vector space)? It just seems that I'd require an infinite number of elements of S to do so. Can anyone help me understand this?
Thanks!
Well, textbooks aren't always correct. Also, 10,000 Frenchmen can still be wrong. Can you tell me what page of what edition of what textbook you're looking at? There might, as I said, be some context that's important here.
[EDIT]: See HallsofIvy's correction below.
No, Ackbeet, the definition of "basis of a vector space" requires that every vector in the space be written as a finite sum of members of the basis because the notion of an infinite sum requires some kind of topology that is not part of the definition of "vector space".
The various "bases" that require infinite sums are all in function spaces which have underlying topologies and are "Hamel bases", not the bases that are defined in linear algebra.
See this thread for a more complete understanding.
It's just the same. You can call it "basis" or "Hamel basis", the same principle applies in both cases: every vector in the space is a unique linear comb. of a FINITE number of elements of the basis (of course, for different vectors there'd be, most probably, different finite elements in the basis a unique lin. comb. of which gives the vector).
The example that's bugging you is a hard one: I can prove a basis must exist for the vectors space of all infinite sequences (say, of real numbers), but I can't point out at one, just as I know there must be a basis (Hamel basis) for the vector space R over the rationals, but I can't point it out in a distinct way.
This is one instance in which an existence theorem and its proof gives us no clue how to actually build the object which existence we can prove. This phenomenum exists in several parts of mathematics.
Tonio
If I remember correctly those are Schauder bases.
You're having trouble with this one, because of what tonio said: An explicit basis is not known. The most I know is that it has to be uncountable!the vector space of infinite-tuples so x = (x1, x2, ...)
I don't know, but many problems posed in vector spaces (for example in numerical analysis) would be greatly simplified if we know a basis (it codifies your whole vector space!)Just curious, if we ever/could (I'm not sure if we even can?) find these Hamel basis for ANY vector space, what kind of consequences would this have?
Well its not really because an explicit basis is not known, but I guess if I'd be able to see one, it would help. I'll try to explain my confusion. Initially, I had thought that the "standard basis" (i.e. B = {(1,0,0,...) (0,1,0,...) and so on}) would be a basis of the vector space I'm considering. But it isn't, because I'd require infinite number of vectors in B in the linear combination for x (where x has an infinite number of nonzero entries). So the reason I find this all confusing is because how is it that there is a basis S (I do know it exists however!) such that a FINITE number of vectors can be used for the linear combination of x which has an INFINITE number of entries. But I suppose that if x = (c1)u1 + (c2)u2 + ... + (cn)un then each ui themselves also have infinite number of entries to "take care of" each entry in x; otherwise, it seems like I'd need an infinite number of u's of S.
See what I mean?