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At first lets introduce some notation.

$\mathcal{A}^n$ is a $n-$dimensional affine space and $V$ is its associated vector space. For any affine subspace of $\mathcal{M}$, its associated vector space it would be denoted as $V_{\mathcal{M}}$.

For the points of the affine space I will use bold capital letters such as $\mathbf{Q},\mathbf{P},\mathbf{R},\mathbf{A},..$ and for the vectors of $V$ bold smaii ones.

The vector $\mathbf{v} \in V$ that produced by $\mathbf{Q}$ and $\mathbf{P}$, it would be denoted as $\mathbf{QP}$

Finally if $S=\left\{\mathbf{O},\mathbf{A_1},\mathbf{A_2},\cdots,\mathbf{A_n}\right\}$ is a coordinate system with origin $\mathbf{O}$ then,

its associated basis of $V$ it would be the $\mathcal{B}_s = \{\mathbf{a_i} = \mathbf{OA_i}: i=1,\cdots,n\}$ and the vector $\mathbf{OX}=\mathbf{x}$, $\mathbf{X} \in \mathcal{A}^n$


Consider a affine transformation $f:\mathcal{A}^n \to \mathcal{A}^n$, with $f(\mathbf{x})=A\mathbf{x}+\mathbf{p}$

What is the general strategy we follow to find the invariant straight lines under the transformation $f$?

Consider a line $\varepsilon$ parallel to $\mathbf{v}$ passing throw point $\mathbf{Q}$, so $$\varepsilon=\left\{ t\mathbf{v}+\mathbf{q}:t \in \mathbb{R} \right\}$$

we have that $$f(\varepsilon)=\left\{ tA\mathbf{v}+A\mathbf{q}+\mathbf{p}:t\in\mathbb{R} \right\}$$

in order to be $\varepsilon=f(\varepsilon)$ it should exists a $\lambda \in \mathbb{R}$ such that $$A\mathbf{v}=\lambda \mathbf{v} \text{ and }A\mathbf{q}=\lambda \mathbf{q}+\mathbf{p}$$

So the problem is, to find the eigenvalues of $A$ and their corresponding eigenvectors and then solve the equation $A\mathbf{q}=\lambda \mathbf{q}+\mathbf{p} $

Theoretically, is everything alright?

Here is a basic example:

Consider a affine transformation $f:\mathcal{A}^2 \to \mathcal{A}^2$, with $f(\mathbf{x})=\begin{bmatrix}2&3\\3 &10\end{bmatrix}\mathbf{x}$, which is the invariant lines of the plane?

using the above method we have that the eigenvalues of $A$ is $\lambda_1=1$ and $\lambda_2=11$ and their corresponding eigenvectors are $\mathbf{v}_1= \begin{bmatrix} -3\\1 \end{bmatrix}$ and $\mathbf{v}_2= \begin{bmatrix} 1\\3 \end{bmatrix}$.

So the invariant lines of the transformation $f$ should have the form $$t\mathbf{v}_i+\mathbf{q}_i \text{ with } i=1,2$$

but how i will find the points $\mathbf{q}_i?$


Edit

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My first try above is wrong but I believe that I make some progress so let me share it with you, I don't delete the former text in order for the reader follow my thoughts.

Consider a affine transformation $f:\mathcal{A}^n \to \mathcal{A}^n$, with $f(\mathbf{x})=A\mathbf{x}+\mathbf{p}$

and a line $\varepsilon$ parallel to $\mathbf{v}$ passing throw point $\mathbf{Q}$, so $$\varepsilon=\left\{ t\mathbf{v}+\mathbf{q}:t \in \mathbb{R} \right\}$$

what is the general strategy we follow to find the invariant straight lines under the transformation $f$?

we have that $$f(\varepsilon)=\left\{ tA\mathbf{v}+A\mathbf{q}+\mathbf{p}:t\in\mathbb{R} \right\}$$

(Here is where my reasoning was wrong)

In order to be $f(\varepsilon)=\varepsilon$ it should be

  1. $f(\varepsilon)\parallel\varepsilon$
  2. and $f(\varepsilon) \cap \varepsilon \neq \varnothing$

So at first we should find for which $\mathcal{v}$ we have that $f(\varepsilon)\parallel\varepsilon$. In other words we should find a $\lambda \in \mathbb{R}$ such that $$A\mathbf{v}=\lambda \mathbf{v}$$

Once we have the find the eigenvalues $\lambda_i$ of $A$ and their corresponding eigenvectors $\mathbf{v_i}$, $i \in I$

We know that every line of the form $$\varepsilon=\left\{ t \mathbf{v_i}+\mathbf{q}:t \in \mathbb{R} \right\}$$ it is mapped throw $f$ to a parallel one.

Now for every $i\in I$ we should the following equation with unknowns the $t_1, t_2$ and $\mathbf{q}$

$$t_1 \mathbf{v_i} + \mathbf{q} = t_2 A \mathbf{v_i} + A \mathbf{q} + \mathbf{p}$$

Even this is a fare general method, the problem seems far from solved. At least in generall.


The general case it seems to be too complicated but we may extract interesting results for the cases $n=2,3$

Many new questions have came to my mind. For example, what happens when the eigenvalues are complex numpers? or when the $\mathbf{p}$ from the last equation belongs to a line parallel to the $f(\varepsilon)$ or..

I will soon add and the solution to the basic example

Feel free to and your thoughts, still I dont know if my reasoning is correct, moreover I am sure somewhere there is a great resource explaining our problem, so if someone know it, please share your reference

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  • $\begingroup$ Nice question. ${}{}$ $\endgroup$ – copper.hat Feb 11 '16 at 16:55
  • $\begingroup$ i edited the post with some new thoughts $\endgroup$ – karhas Feb 12 '16 at 3:17
  • $\begingroup$ If you work in homogeneous coordinates, your problem becomes that of finding the two-dimensional invariant subspaces of $M=\small{\begin{bmatrix}A&\mathbf p\\\mathbf 0^T&1\end{bmatrix}}.$ This is fairly easy if $M$ is diagonalizable (or its minimal polynomial factors into linear and irreducible quadratic factors), but rather challenging when there are defective eigenvalues. $\endgroup$ – amd Apr 20 at 7:21

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