Proper invariant subspace given minimal polynomial

Question

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!

Answer 1

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.