Vectorialization of Affine Space is Vector Space

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

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

Let $\struct {\mathcal E, +, \cdot}$ be the vectorialization of $\mathcal E$.

Then $\struct {\mathcal E, +, \cdot}$ is a vector space.


By the definition of the vectorialization of an affine space, the mapping $\Theta_\mathcal R : k^n \to \mathcal E$ defined by:

$\displaystyle \map {\Theta_\mathcal R} {\lambda_1, \ldots, \lambda_n} = p_0 + \sum_{i \mathop = 1}^n \lambda_i e_i$

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 if and only if its inverse ${\Theta_\mathcal R}^{-1}$ is a linear transformation.

Therefore, it suffices to show that:

$\forall p, q \in \mathcal E, \mu \in k: \map { {\Theta_\mathcal R}^{-1} } {\mu \cdot p + q} = \mu \cdot \map { {\Theta_\mathcal R}^{-1} } p + \map { {\Theta_\mathcal R}^{-1} } g$


\(\displaystyle \map { {\Theta_\mathcal R}^{-1} } {\mu \cdot p + q}\) \(=\) \(\displaystyle \map { {\Theta_\mathcal R}^{-1} } {\map {\Theta_\mathcal R} {\mu \cdot \map { {\Theta_\mathcal R}^{-1} } p} + q}\) Definition of $\mu \cdot p$
\(\displaystyle \) \(=\) \(\displaystyle {\Theta_\mathcal R}^{-1} \map {\Theta_\mathcal R} {\mu \cdot \map { {\Theta_\mathcal R}^{-1} } p + \map { {\Theta_\mathcal R}^{-1} } q}\) Definition of $+$ in $\mathcal E$
\(\displaystyle \) \(=\) \(\displaystyle \mu \cdot \map { {\Theta_\mathcal R}^{-1} } p + \map { {\Theta_\mathcal R}^{-1} } q\) because $\map { {\Theta_\mathcal R}^{-1} } {\Theta_\mathcal R}$ is the identity mapping

This is the required identity.