Let $V$ be a finite dimensional vector space and $T:V\rightarrow V$ be a linear map.
Suppose $U$ is a T-invariant subspace. Define $\overline{T}:V/U \rightarrow V/U$ by $\overline{T}(v+U)=T(v)+U$.
Suppose also that $U$ has no non-zero proper T-invariant subspaces, and $V/U$ has no non-zero proper $\overline{T}$-invariant subspaces, is $U$ the only non-zero proper T-invariant subspace of $V$?
I thought no with this example in mind: let $V$ be $\mathbb{R}^3$ and T be a rotation along the z-axis. Let $U$ be the x-y plane, which is T-invariant, then the z-axis is T-invariant. I'm not sure if $V/U$ in this case can be seen as the z-axis (plus the x-y plane).
If this idea doesn't work, could you also explain how I could solve this? Thank you!
Your example actually satisfies the condition which as you suspect is not true. Here is a counterexample. Consider the linear map on $\mathbb{R}^3$ with matrix representation
$$ \begin{pmatrix} 1&0&0\\ 0&1&0\\ 0&0&3 \end{pmatrix}$$ and the subspace $U=span\{e_1,e_3\}$ which is $T$ invariant. The space $\mathbb{R}^3/U$ is one dimensional with basis $\{\overline{e_2}\}$. Although it is $\overline{T}$ invariant, there are no proper $\overline{T}$ invariant subspaces (since it's one dimensional). However, within $U$ the spaces generated by $\{e_1\}$ and $\{e_3\}$ are $T$ invariant which contrary to the proposition.