Vectorialization of Affine Space is Vector Space

Theorem

Let $\EE$ be an affine space over a field $K$ with difference space $E$.

Let $\RR = \tuple {p_0, e_1, \ldots, e_n}$ be an affine frame in $\EE$.

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

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

Proof

By the definition of the vectorialization of an affine space, the mapping $\Theta_\RR : K^n \to \EE$ defined by:

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

is a bijection from $K^n$ to $\EE$.

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

By General Linear Group is Group:

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

Therefore, it suffices to show that:

$\forall p, q \in \EE, \mu \in K: \map { {\Theta_\RR}^{-1} } {\mu \cdot p + q} = \mu \cdot \map { {\Theta_\RR}^{-1} } p + \map { {\Theta_\RR}^{-1} } g$

Thus:

\(\ds \map { {\Theta_\RR}^{-1} } {\mu \cdot p + q}\)

\(=\)

\(\ds \map { {\Theta_\RR}^{-1} } {\map {\Theta_\RR} {\mu \cdot \map { {\Theta_\RR}^{-1} } p} + q}\)

Definition of $\mu \cdot p$

\(\ds \)

\(=\)

\(\ds {\Theta_\RR}^{-1} \map {\Theta_\RR} {\mu \cdot \map { {\Theta_\RR}^{-1} } p + \map { {\Theta_\RR}^{-1} } q}\)

Definition of $+$ in $\EE$

\(\ds \)

\(=\)

\(\ds \mu \cdot \map { {\Theta_\RR}^{-1} } p + \map { {\Theta_\RR}^{-1} } q\)

because $\map { {\Theta_\RR}^{-1} } {\Theta_\RR}$ is the identity mapping

This is the required identity.

$\blacksquare$