I'm reading Linear Algebra book, stating that points are elements in the affine space $\mathcal{A}$, associated to a certain linear space $\mathbb{A}$. The elements in $\mathbb{A}$ are free vectors. There's function $\phi:\mathcal{A}\times \mathcal{A} \rightarrow \mathbb{A}$ satisfying the following properties:
- Given any point $A$ and free vector $u$, then there is one and only one point $B$ such that $\phi(A,B)=u$.
- For any points $A,B,C$ we have $\phi(A,B)+\phi(B,C)=\phi(A,C)$
There's a bijection between $\mathcal{A}$ and $\mathbb{R}^3$, and another between $\mathbb{A}$ and $\mathbb{R}^3$.
Given points $A,B$, and a vector $u$, we can write $A+u=B$ as notation for $\phi(A,B)=u$ (the sum between a point and a vector).
Using this notation it's natural to define a straight line as the set of points of form $A+\lambda u\ \forall_{\lambda \in \mathbb{R}}$. This is a subset of the affine space.
Using the bijections above we can write the straight line as $(x_1,x_2,x_3)=(a_1,a_2,a_3)+\lambda (u_1,u_2,u_3)$ (using the coordinates for the basis we fixed in $\mathbb{A}$ and $\mathcal{A}$)
I'm not sure why we can do $(x_1,x_2,x_3)=(a_1+\lambda u_1,a_2+\lambda u_2,a_3+\lambda u_3)$ and get a point… When we add ($+$ in $\mathbb{R}^3$) those components, we're adding representations of a free vector and a point, which belong to different sets…how are we sure that it respects the definition used above for a sum between a point and a vector?
Best Answer
Normally when we pick coordinates in linear algebra, we pick compatible choices on all types of things involved.
Name the maps $i : \mathcal{A} \to \mathbb{R}^3$ and $j : \mathbb{A} \to \mathbb{R}^3$
A typical presentation of this would, after having selected some linear isomorphism $j$, pick some point $O$ in the affine space to be mapped to the origin and then define $i$ to be the map given by $i(P) = j(\phi(O,P))$.
Then, you do have that $i(P + v) = j(\phi(O, P+v)) = j(\phi(O,P) + v) = i(P) + j(v)$
(P.S. it's convenient to use the notation $Q-P$ for $\phi(P,Q)$)
If you had a different $i$ that wasn't (equivalent to) something constructed this way, then you wouldn't be able to compute the affine space operations via vector operations on their coordinate representations... but you shouldn't have picked an $i$ that was incompatible with your $j$.
It's somewhat more typical, I think, to do the construction with just $\mathcal{A}$ and $\mathbb{A}$; that is, once you pick an element $O \in \mathcal{A}$, you define a bijection $h : \mathcal{A} \to \mathbb{A}$ by $h(P) = \phi(O,P)$.
(then, with everything in terms of $\mathbb{A}$, you could then consider a map $\mathbb{A} \to \mathbb{R}^3$ if you are inclined to do so)