TITLE: Minimal polynomial and invariant subspaces.
QUESTION [8 upvotes]: Let $T$ be a linear operator on the $n$ dimensional vector space $\mathbb{V}$. Suppose that $\mathbb{V} = \sum_{i=1}^{k}W_i$ where each $W_i$ is $T$ - invariant. Let $\mu_{T_i}$ be the minimal polynomial of the operator restricted to $W_i$. If the minimal polynomials of all restrictions are coprime, that is if $\gcd(\mu_{T_i}, \mu_{T_j}) = 1$ for $i\neq j$, is it true that the subspaces $W_i$ are independent? That is, $\mathbb{V} = \bigoplus_{i=1}^k W_i$. Perhaps more generally, is it true that if the minimal polynomials of two subspaces are coprime, then the two subspaces are independent? 
The converse is easily seen to be false, the identity operator is a clear counterexample. I haven't been able to find an easy counterexample to the above statement however. A proof or counterexample would be appreciated.

REPLY [2 votes]: Here is one more variant. Due to the relative primality, $\mu_i[T]$ acts invertibly on each subspace $W_j$ with $j\neq i$: if $r,s\in K[T]$ are Bezout coefficients for $\mu_i,\mu_j$, in other words $r\mu_i+s\mu_j=1$, then restricted to$~W_j$ the operator $r[T]\mu_i[T]$ acts as the identity. This is the kind of situation that easily give directness of the sum of the subspaces $W_i$: given a hypothetical non-trivial relation $0=w_1+\cdots+w_k$ with $w_i\in W_i$ for all$~i$, take one for which a minimal number of nonzero vectors among the $w_i$ (but of course at least one; in fact there now clearly must be at least two nonzero among them); with $w_i\neq0$, apply $\mu_i[T]$ to the relation to find a new relation with exactly one less nonzero vector, contradicting minimality.