Can every basis of a vector space be reduced to the standard basis? 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 $\mathbb{R}^n$?
 A: 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.
A: 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.
