Following the linear-algebraic approach to projective spaces, I've come across this exercise, useful to prove a notable isomorphism of the group PGL. I don't enter into details of this circumstance since they are not essential to understand the question.
Given a vector space V, consider in $V-\{0\}$ the equivalence relation $E$ in which two vectors are equivalent if and only if they are proportional, i.e. one is a scalar multiple of the other. Call $p$ the map associating to every non-zero vector its $E$-equivalence class. Prove that, given two automorphism $f$ and $g$ of $V$, it is true that:
for every non-zero vector $v$, $p_f(v) = p_g(v)$
if and only if
there exists a non-zero scalar $r$ such that, for every non-zero vector $v$, $g$ maps $v$ into $rf(v)$ (i.e. $g(v)=rf(v)$).
By now, I can only prove that, fixed a vector v, there exists a non-zero scalar r such that, $g(v)=rf(v)$. I'd like to prove that for different vectors one gets the same $r$.
Thank you.
There are different approaches one can take here, but I think that the following is the most straightforward.
Hint: Show that if $v_1,v_2$ are (linearly independent vectors) such that $g(v_1) = r_1f(v_1)$ and $g(v_2) = r_2f(v_2)$ with $r_1 \neq r_2$, then $g(v_1 + v_2)$ cannot be a multiple of $f(v_1 + v_2)$.
As a further hint, a key point here is that $f$ being an automorphism implies that $f(v_1),f(v_2)$ are linearly independent.
An alternative approach that you might find informative: note that $p_f = p_g$ if and only if $p_{f^{-1}f} = p_{f^{-1}g}$. Thus, we may reduce our consideration to the case where $f = \text{id}_V$.
Now if $f = \text{id}_V$, note that the statement "there exists a non-zero scalar $r$ such that, for every non-zero vector $v$, $g$ maps $v$ into $rf(v)$ (i.e. $g(v)=rf(v)$)" is a statement about the eigenvalues and eigenvectors of $g$.