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How can it happen to find infinite bases in $\mathbb R^n$ if $\mathbb R^n$ does not admit more than $n$ linearly independent vectors?

Also considered that each basis of $\mathbb R^n$ has the same number $n$ of vectors.

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Why do you think it can for $n$ finite? –  Ross Millikan Nov 30 '12 at 18:16
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It cannot happen. –  Andrea Mori Nov 30 '12 at 18:43
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Maybe there is some confusion here about the finite dimensional vector space $\mathbb{R}^n$ and some infinite dimensional vector space of functions with domain and range in some subset of $\mathbb{R}^n$. –  Peter Sheldrick Nov 30 '12 at 21:25
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I don't understand what you mean by "infinite bases." –  Qiaochu Yuan Nov 30 '12 at 21:34
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Infinite basis over $\mathbb R$ or over $\mathbb Q$ ? ;) –  N. S. Nov 30 '12 at 21:36

5 Answers 5

There is no infinite basis of $\mathbb R^n$. Maybe someone intended to say there are infinitely many bases, but clumsily expressed it by saying there are "infinite bases". I see this particular misuse of terminology frequently.

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shouldn't there be just one basis for R-enth then? The one made up of the only n linearly independent vectors allowed in R-enth? –  Matteo Nov 30 '12 at 18:22
    
anyway, I quote precisely the text: "In R-enth there exist infinite basis." –  Matteo Nov 30 '12 at 18:23
    
The set $\{(1,0),\ (0,1)\}$ is a basis of $\mathbb R^2$. The set $\{(1,-1),\ (1,1)\}$ is another basis of $\mathbb R^2$. And there are infinitely many others. –  Michael Hardy Nov 30 '12 at 18:24
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@Matteo : Are you translating the text into English from some other language? Maybe it's a mistranslation to say there are infinite bases, and it ought to by translated by saying "There are infinitely many bases." –  Michael Hardy Nov 30 '12 at 18:38
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@Michael: that could easily be the case. I can vouch for the case with Spanish, where "infinitely many bases" would be "infinitas bases" while "infinite basis" would be "base infinita". –  Martin Argerami Nov 30 '12 at 18:59

There are infinitely many bases (plural) of $\mathbb{R}^n$, but each basis (singular) must contain $n$ (finite) linearly independent vectors, if it is in fact a basis.

  • Take any basis $B_0 = \{\vec b_1, \vec b_2, ..., \vec b_n\}$ for $\mathbb{R}^n$.

  • Take any scalar $c_i\neq 0$ (there are infinitely many such $c_i$, $i\in \mathbb{N}$).

  • Then there infinitely many unique bases, $B_i =\{c_i\vec b_1, \vec b_2, ... , \vec b_n\}$ for $\mathbb{R}_n$, with each basis $B_i$ corresponding to a particular (unique) scalar $c_i$

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It can happen, if you consider $\mathbb R^n$ as a vector space over the field $\mathbb Q$. Then you certainly have infinite dimension and you can construct define the Hamel basis.

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Related: mathoverflow.net/questions/46063/… –  tohecz Nov 30 '12 at 22:40

Let $E=\{e_1,...,e_n\}$ be the standard basis in $\mathbb{R}^n$.

For each $\lambda\neq 0$, let $E_\lambda = \{\lambda e_1, e_2,...,e_n\}$. Then each $E_\lambda$ is a distinct basis of $\mathbb{R}^n$.

However, each $E_\lambda$ has exactly $n$ elements.

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There are infinitely many finite bases for $\Bbb R^n$, simply because we can always replace one vector by a scalar multiple of itself. There are infinitely many scalars, so there are infinitely many bases.

Note that for $n=2$ a subspace is simply a straight line going through $(0,0)$. There are infinitely many of those simply because we can just turn the lines through infinitely many angles. Similarly for higher dimensions, you have lines, planes, and so on, all of which you can twist and shake around to get an infinite number of subspaces.

If however the underlying field was a finite field then indeed a finitely dimensional space could have only a finite number of subspaces and a finite number of bases. In fact the whole space was finite!

While a finite dimensional space [over $\Bbb R$] is an object which is "essentially finite" (because a lot of its structure is determined by the behavior over a finite set) it is still infinite as a set.

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