# Proper invariant subspace given minimal polynomial

I am wondering about the following linear algebra problem:

Let the minimal polynomial of $$T$$ on a finite-dimensional vector space $$V$$ be $$p^2$$, where $$p$$ is irreducible. Is it true that $$V$$ contains a proper $$T$$ invariant subspace?

Is there a relationship between the irreducibility of factors showing up in the minimal polynomial and possible invariant subspaces?

I did find the following: Minimal polynomial is irreducible if the only $T$-invariant subspaces of $V$ are $V$ and $0$

Is there some statement about the converse? Furthermore, I'm not exactly dealing with the converse here, since the minimal polynomial is reducible. I guess, what I can say is, the minimal polynomial splits as a product of irreducible factors (not necessarily linear).

Thanks!

Define $$W=p(T)(V) = \{p(T)(v)\colon c\in V\}.$$ Note that $$W\ne\{0\}$$, since that would imply that the minimal polynomial of $$T$$ divides $$p$$. But note that $$p(T)(W) = p^2(T)(V)=\{0\}$$. This implies that $$W\ne V$$, since otherwise $$p(T)(W)=p(T)(V)=W\ne\{0\}$$. Therefore $$W$$ is a proper subspace of $$V$$. Finally, for every $$w\in W$$, we have $$w=p(T)(v)$$ for some $$v\in V$$, and so $$T(w) = T(p(T)(v)) = (Tp(T))(v) = p(T)(Tv) \in W,$$ and so $$W$$ is $$T$$-invariant.