Ytukyg

Suppose we have affine transformation:

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

How can we find the equations of all invariant lines?

Thanks in advance

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 $$f(a,b)=a\g(a,b)=b$$

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

$$
p = p_0 + lambda vec v
$$

then

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

or

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

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.

Working in homogeneous coordinates, the point transformation $mathbf x'=Mmathbf x$ 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 $mathbf ltimes M^{-T}mathbf l=0$.

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 $(0,0,1)^T$.