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Consider the $n$-dimensional vector spaces $V\in\mathbb{R}^n$. Given a non-standard basis $B$ for the vector space $V$, can it be reduced to the standard basis on $\mathbb{R}^n$?

Cameron Buie
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    What do you mean by "reduce to the standard basis"? – Daniel Fischer Aug 09 '13 at 15:54
  • @CameronBuie Sometimes a question comes from a conceptual mistake, and I think that's the case here - we are left with a question that still has conceptual mistakes, but hides what might be the key to unlocking the mistake. – Thomas Andrews Aug 09 '13 at 16:18
  • @Thomas: Fair point. Rolled back. Hopefully the OP clarifies the question. – Cameron Buie Aug 09 '13 at 16:30
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    Chase, did you intend to say that $V\in\Bbb R^n$? This means that $V$ is an element of the canonical $n$-dimensional real vector space. It does not mean that $V$ is itself an $n$-dimensional real vector space. If you could clear this up, and answer Daniel's question, I think it would be a great deal easier to answer your question. – Cameron Buie Aug 09 '13 at 16:32

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Any two bases of a vector space are equivalent, in the sense that there exists a linear transformation will map one basis to the other in a 1-to-1 mapping.

In fact one characterization of a basis is that any mapping of the basis to arbitrary vectors in a vector space determines a linear transformation. So take one basis and assign distinct "standard basis vectors" to each element of that (non-standard) basis, and then this assignment extends by linearity to a linear transformation on the vector space as a whole.

hardmath
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In this constellation $V$ and $\mathbb{R}^n$ are even isomorphic, because there is a uniquely determined linear map $f:V\to\mathbb{R}^n$, which has to be bijective due to the fact that we know a basis for both vector spaces - hence we can define a mapping $b_i\mapsto e_i$ for every $b_i\in B$, where $e_i$ is the $i-$th vector of the standard basis of $\mathbb{R}^n$.

Geometrically speaking it is only reasonable to say that $V$ in terms of $B$ can be "reduced" to $\mathbb{R}^n$ in terms of its standard basis, if $V=\mathbb{R}^n$, in the sense of $V$ (in the basis $B$) beeing an "odd version" version of $\mathbb{R}^n$ (in the standard basis). In any other case the two vector spaces have nothing in common than abstract properties. Think of $V$ beeing the vector space of polynomials. We know that, if $V$ and $\mathbb{R}^n$ have the same dimension, they "must behave the same" (: be isomorphic) because of the existence of $f$; but it is impossible to say anything else.

Ruben
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