Prove that $f \circ f$ is homothety or a constant transformation

Question

I am required to prove that, if $X$ is an 2-dimensional affine space and $f:X\rightarrow\ X$ is an affine transformation such that the $Tr([f`]_b)=0$, prove that $f\circ f$ is homothety or a constant transformation. If the trace is $0$, then the matrix $[f`]_b$ has the determinant of the form $-a^{2}-bc$, where $a, b$ are the real numbers on the first row and $c, -a$ the real numbers on the second row. I think that after computing the values in the equation of the affine transformation, the scale factor is $a^2+bc$. However, I couldn`t progress in a full proof. Any help please?

Answer 1

The affine transformation $f : X \to X$ is written, in coordinates:

$$f(x) = Ax+B, $$

where $A$ is a $2\times 2$ matrix. As the trace $\text{Tr}(A)=0$, we can write

$$ A=\left[\matrix{a & b \\c & -a}\right]. $$

The transformation $f^2=f\circ f$ is then written

$$ f^2(x)=f(f(x))=f(Ax+B)=A(Ax+B)+B=A^2x+AB+B. $$

Hence the linear part $(f^2)'$ of $f^2$ is represented by the matrix

$$ A^2 = \left[\matrix{a^2+bc & 0 \\ 0 & cb+a^2}\right] = (a^2+bc)\,I_2$$

and therefore we can write

$$ f^2(x)=(a^2+bc)\,I_2x+AB+B = (a^2+bc)x+AB+B. $$

We conclude that $f^2$ is a homothety having scalar factor $\lambda=a^2+bc$, and that it is a constant transformation if and only if $\lambda=a^2+bc=0$.

$\bf \text{Note.}\quad$ We observe that $\det(A)=-a^2-bc=-\lambda.$ Therefore, $f^2$ is constant if and only if $\det(A)=\det(f')=0$.