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, we know that the map $\Theta_{\mathcal R} : k^n \to \mathcal E$ 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 $k^n \to \mathcal E$.

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 we know that $\Theta_{\mathcal R}$ is a linear transformation if and only if $\Theta_{\mathcal R}^{-1}$ is a linear transformation.

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

We find that

This is the required identity.