Vectorialization of Affine Space is Vector Space

Theorem
Let $\mathcal E$ be an affine space over a field $k$ with difference space $E$.

Let $\mathcal R = \left({p_0, e_1, \ldots, e_n}\right)$ be an affine frame in $\mathcal E$.

Let $\left({\mathcal E, +, \cdot}\right)$ be the vectorialization of $\mathcal E$.

Then $\left({\mathcal E, +, \cdot}\right)$ is a vector space.

Proof
By the definition of the vectorialization of an affine space, the mapping $\Theta_\mathcal R : k^n \to \mathcal E$ defined by:
 * $\displaystyle \Theta_\mathcal R \left({\lambda_1, \ldots, \lambda_n}\right) = p_0 + \sum_{i \mathop = 1}^n \lambda_i e_i$

is a bijection from $k^n$ to $\mathcal E$.

Therefore, by Homomorphic Image of Vector Space, it suffices to prove that $\Theta_\mathcal R$ is a linear transformation.

By General Linear Group is Group:
 * $\Theta_\mathcal R$ is a linear transformation its inverse ${\Theta_\mathcal R}^{-1}$ is a linear transformation.

Therefore, it suffices to show that:
 * $\forall p, q \in \mathcal E, \mu \in k: {\Theta_\mathcal R}^{-1} \left({\mu \cdot p + q}\right) = \mu \cdot {\Theta_\mathcal R}^{-1} \left({p}\right) + {\Theta_\mathcal R}^{-1} \left({q}\right)$

Thus:

This is the required identity.