Let $V$ be a finite-dimensional vector space over a field $\mathbb{F}$ and let $T$ be a linear operator on $V$. Show that $T$ is indecomposable if and only if there is a unique maximal proper $T$-invariant subspace of $V$.
I have came up with a proof of the forward direction but I’m having trouble with the converse. This problem is from Cooperstein’s Advanced Linear Algebra and the hint for this problem says to prove the contrapositive for the converse. That is, if $T$ is decomposable then there exists more than one maximal proper $T$-invariant subspace.
I know that there’s a theorem that says that $T$ is indecomposable if and only if $T$ is cyclic and it’s minimal polynomial is the power of an irreducible polynomial.
So, I assumed that $T$ is decomposable which means that there exists non-trivial $T$-invariant subspaces of $V$, call them $U$ and $W$, such that $V = U \oplus W$. From here, I tried to prove that $U$ and $W$ were both maximal but that didn’t lead anywhere as far as I was aware. In my attempt to do that, I assumed that there was some $T$-invariant subspace $X$ that properly contains $W$ and I tried to show that $X$ had to be all of $V$ but I wasn’t sure how to go about this. Moreover, this might not be a reasonable to go about this.
A hint, different idea to try, or a solution would be much appreciated.
Extend $U$ to a maximal proper invariant subspace $U'$, and find a corresponding $W'$ for $W$. Since the maximal proper invariant subspace is unique, we have $U' = W'$, containing both $U$ and $W$. Is this possible?