I'm studying Computer Vision and my lecturer stated that:
The affine transformation maps "line at infinity" to "line at infinity".
I'm trying to prove it as part of my preparation for the final exam. My lecturer suggested to check the points on the line for that. Also in the lecture, it was explained that "line at infinity" is a line of form $I_{\infty}=\begin{bmatrix}0 & 0 & 1\end{bmatrix}^{T}$ (where $T$ is transpose). So for any ideal point (has $0$ in the third coordinate) $p=\begin{bmatrix}x & y & 0\end{bmatrix}^{T}$: $$ I_{\infty}^{T}p=\begin{bmatrix}0 & 0 & 1\end{bmatrix}^{T}\begin{bmatrix}x & y & 0\end{bmatrix}^{T}=0 $$ It was in the context of projective spaces.
How can I prove the above theorem of Affine transformation that maps "line at infinity" to "line at infinity"? Tried to look online but only found a statement here (page 5).
Also, I guess that the definition of "Ideal points" and "line at infinity" are much more broad then the one I got from the lecture, but since I need to stick to those definitions, how one can prove the above statement? (It's Computer Vision course after all)
Mo Pol Bol's comment is comprehensive and effective. Here I just want to highlight the theory behind that comment, if that can shed more light on the situation in general.
Let ${\cal A}(E)$ be an affine space over the $K$-vector space $E$ (where $K$ is a field).
The canonical projective extension of ${\cal A}(E)$ is the projective space ${\cal P} (E\times K)$. The function
$$ J = \left\{ \begin{matrix} {\cal A}(E) & \longrightarrow & {\cal P}(E\times K)\\[1ex] P & \longmapsto & [P-O,1] \end{matrix} \right. $$ where $O$ is any fixed point of ${\cal A}(E)$, and $[P-O,1]$ is an abbreviation for $\langle(P-O,1)\rangle$, is an immersion of ${\cal A}(E)$ into ${\cal P}(E\times K)$. The set
$$ l_\infty := {\cal P}(E\times K) \setminus J({\cal A}(E)) $$ is an hyperplane of ${\cal P}(E\times K)$, called the improper hyperplane or the hyperplane at infinity of ${\cal A}(E)$. Obviously it is a line if the affine space is an affine plane.
It is easy to see that
$$ l_\infty = \{[a,0]\;\;:\;\;a\in E, \;a\neq 0\}. $$
Each affinity $f : {\cal A}(E) \longrightarrow {\cal A}(E)$ extends to a projectivity $\tilde f :{\cal P}(E\times K) \longrightarrow {\cal P}(E\times K)$:
$$ \tilde f([a,t]) = [df(a), t] \qquad\qquad \big(a\in E,\;t\in K,\; (a,t)\neq(0,0)\big), $$
where $df$ is the linear part of $f$. $\big[$Remember that a function $f : {\cal A}(E) \rightarrow {\cal A}(E)$ is affine iff there is a linear function $L : E\rightarrow E$ such that $f(P)-f(Q)=L(P-Q)$ for all $P,Q\in {\cal A}(E)$. We denote $L$ by $df.\big]$
Now the conclusion is almost obvious:
$$ \tilde f([a,0]) = [df(a),0]\qquad\qquad \forall\,a\neq0 $$
so
$$ \tilde f(l_\infty) = l_\infty $$
because $df$ is bijective, being $f$ an affinity.