Let $A$ be a $n$-dimensional affine space ($ 2 \leq n$) and let $\Phi:A\to A$ be a bijective affine mapping, which isn't the identity mapping, with the following property:
For all points $p$ and $q$ in $A$ such that $p \neq \Phi(p)$ and $q \neq \Phi(q)$ we have that the line connecting $p$ and $\Phi(p)$ is parallel to the line connecting $q$ and $\Phi(q)$.
Prove that if $\Phi$ has a fixed point, then the set of all fixed points is a hyperplane.
My attempt:
$\Phi$ is a mapping of the form $\Phi:x \mapsto Mx + b$ where $M$ is an invertible matrix and $b$ some translation vector. By a change of coordinates, we may choose the fixed point of $\Phi$ as our origin. Then $\Phi$ takes the form of a linear mapping $\Phi:x \mapsto Mx$. So then $x$ is a fixed point of $\Phi \iff Mx = x \iff (M-I)x=0 \iff x \in \ker(M-I) \iff x$ is an eigenvector of $M$ corresponding to the eigenvalue $1$. The claim then is equivalent to the claim that the eigenspace of $M$ for the eigenvalue $1$ has dimension $n-1$.
I don't know how to proceed or if I'm on the right track. I appreciate any hints or alternative soltuions.
Thank you.
Firstly, I would like to notice that $\Phi$ can be the identical map, for which The set of fixed points is not a hyperplane. So I think that your exercise is in some sense incomplete.
However, supposing $\Phi \neq id_A$ gives us no problems. So, let's proceed.
You didn't use your hypothesis on $\Phi$. It is equivalent on saying that for all points $v,w$ you have that $v- \Phi (v), w- \Phi (w)$ are linearly dependent.
Notice that, from $\Phi \neq id_A$, you have that $M-I \neq 0$, so $\dim \ker (M-I) \leq n-1$ . We have to prove that in fact equality holds.
Now, from your hypothesis you have that $\dim \mbox{ Im } (M-I) \leq 1$.
And now $$ n= \dim \ker (M-I) + \dim \mbox{ Im } (M-I) \leq n-1 + 1 = n$$ so equality must hold.