Re: What is a basis for vector space of {(a_1,a_2,...)} a_i real?
From: N. Silver (mathelp_at_worldnet.att.net)
Date: 08/30/04
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Date: Mon, 30 Aug 2004 07:38:34 GMT
Stephen J. Herschkorn wrote:
> Nat wrote:
> >The standard basis is:
> >e1 = (1,0,0,...),
> >e2 = (0,1,0,...),
> >e3 = (0,0,1,...),...
I had in mind an infinite dimensional space.
Its basis would be infinite,{e1, e2, e3,...}.
> >Then a vector (a_1,a_2,a_3,...) = (a_1)e1 + (a_2)e2 +...
> >
> >This space is isomorphic to the space of polynomials,
> >however you want to choose the coefficients (either only
> >rationals or real coefficients). The standard basis for this
> >space is {1, x, x^2, x^3,...}.
> Wrong. How would you express the vector (1,1,1,...) as a finite linear
> combination of elements of this "standard basis"?
I never thought of the basis as being finite. However, I'm pretty sure the
poster's question is beyond my current understanding, since Wade has
mentioned the term "Hamel basis," which I have heard of but am not
familiar with.
> By definition, a subset B of a vector space V over a field F iff
> for every v in V, there exist *finite* subsets {f1, f2,.., f_n} of
> F and {b1, b2,..., b_n} of B such that v = sum (i = 1..n, f_i b_i).
> Correction: I meant to write that such a set B is a *basis* for V
iff...
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