Find all invariant lines of an affine transformation in plane

Question

Suppose we have affine transformation:

$$\begin{align}x^* &= 2x+y-2\\ y^* &= -3x- y\end{align}$$

How can we find the equations of all invariant lines?

Thanks in advance

Answer 1

The equation of a line originally is $y=ax+b$. After transformation is $y^*=a^*x^*+b^*$. Plug in your equations for $x^*$ and $y^*$ into this equation, group the terms so you can rewrite it as $$y=f(a^*,b^*)x+g(a^*,b^*)$$ Since the line is invariant, you have the same coefficients so $$f(a,b)=a\\g(a,b)=b$$

Answer 2

Calling $p = (x,y),\, b = (-2,0)$ we have the generic line as

$$ p = p_0 + \lambda \vec v $$

then

$$ p_0 + \lambda \vec v = A( p_0+\lambda\vec v) + b $$

or

$$ A p_0-p_0+b = \lambda(\vec v-A\vec v),\,\,\forall \lambda\in \mathbb{R} $$

so we have

$$ (A-I_2)\vec v = 0\\ (A-I_2)p_0 + b = 0 $$

now solving for $\vec v, p_0$ we have the invariant lines.

Answer 3

Working in homogeneous coordinates, the point transformation $\mathbf x'=M\mathbf x$ transforms lines as $\mathbf l'=M^{-T}\mathbf l$. The invariant lines $\mathbf l=(\lambda,\mu,\tau)^T$ are those for which $M^{-T}\mathbf l$ is a nonzero multiple of $\mathbf l$, so finding them amounts to finding eigenvectors of $M^{-T}$ that correspond to real nonzero eigenvalues. Alternatively, since the homogeneous coordinates are three-dimensional, the invariant lines are the solutions of $\mathbf l\times M^{-T}\mathbf l=0$.

Note that the line at infinity is always an invariant line of an affine transformation: the last column of $M^{-T}$ for such a transformation is $(0,0,1)^T$.