This question has already been asked here, but I have some more questions to ask.
The usual answer to solve this problem is to assume $T$ is invertible and notice that the characteristic polynomial has a real root. I was wondering if there is a $L$ of the form $\{ v_1 + kv_2 \mid k\in \mathbb{R} \}$ where $v_1, v_2 \in \mathbb{R}^3$ and $v_2 \ne 0$ s. t. $T(L)=L$ without assuming invertibility of $T$.
Is it possible to find such a $L$? I tried hard, but couldn't get far.
If, for instance, $T=0$, then this becomes very difficult, as $T$ maps everything to the origin, and there is no line in the range of $T$. You either need $T$ invertible, or relax the problem to $T(L)\subseteq L$. (There are more "accurate" restrictions you could place too, but they become either more contrived, or basically saying directly that there is a line, so there is no problem left.)
The request is that there exist $u$ and $v\ne0$ such that, for every $k$, $T(u+kv)=u+hv$, for some $h$, and the restriction of $T$ to the line doesn't map everything to $u$.
In particular, $T(u)=u+av$ and $T(u+v)=u+bv$, so $T(v)=u+bv-(u+av)=(b-a)v$. Hence $b-a$ must be a real eigenvalue. It must also be nonzero, otherwise the line would be mapped to $u$. So the matrix needs to have a nonzero real eigenvalue, which need not be the case if the rank of $T$ is $2$.
