Change of basis with a nonlinear operator Given a vector space $V$ and its two basis: $\mathcal{B}$ given by vectors $\{e_i\}$ and $\mathcal{B}'$ given by vectors $\{e'_i\}$, why are the two basis necessarily connected by a linear transformation $$e'_i = \Lambda_{ij}e_j?$$ Why do we exclude the possibility of a nonlinear transformation, e.g. $$e'_i = \Lambda_{ijk}e_j e_k?$$
 A: I will try to give an example of why this isn't terribly useful, even in the case where the nonlinear transformation is constrained to be nonsingular and transforming a basis to a basis, although it's probably clear that in the general case, every nonlinear transformation will probably break the linear independence of some of the bases.
Let us take a one-dimensional vector space $V$ with a vector $a$, a basis $B=\{b\}$ and another $B'=\{b'\}$. 
Let us also take a non-linear function $f: V \to V$ acting in the following way: $f(x) = (\text{Rep}_B(x))^3 b$, clearly, $f$ is bijective so at least when applied to all vectors in $V$ one gets another sets that spans $V$.
For contrast, Let $l$ be some linear bijection, $l: V \to V$.
let $y=l(a)$. Now $y$ is related to $a$, and this relation transfers nicely to their coordinates: $y=l(a)=l(a_1 b) = a_1 l(b)$. 
If $b'=l(b)$, then we can use $y=a_1 b'$.
The non-usefulness comes from the fact that the coordinates 'break', become 'useless', when we consider the non-linear function.
$z=f(a)=f(a_1 b) \neq a_1 f(b)$.
If $b'=f(b)$, it gives us nothing we can use, since $z \neq a_1 b'$. 
In short a non-linear change of basis takes us out of linear algebra and the crucial identification between vector and coordinates becomes unworkable in the linear algebraic context.
A: Is possible but isn't terribly useful. The important fact is that always exists a linear transformation between two basis. Also, a nonsingular linear transformation always transforms a basis in a basis. With nonlinear transformations this fails.
