Minimal polynomial and invariant subspaces.

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.

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Yes it is true. Here is a sketch of a proof by contraposition.

If $W:=W_j\cap \sum\limits_{i\neq j}W_i\neq\{0\}$ for some $j$, then $W$ is a nonzero invariant subspace for $T$. The minimial polynomial $p$ of the restriction of $T$ to $W$ must divide both $\mu_{T_j}$ and the minimal polynomial $q$ of $T$ restricted to $\sum\limits_{i\neq j}W_i$. But $q$ divides $\prod\limits_{i\neq j}\mu_{T_I}$, and this implies that each irreducible factor of $p$ must divide both $\mu_{T_j}$ and $\mu_{T_i}$ for some $i\neq j$. Therefore $\gcd(\mu_{T_i}, \mu_{T_j}) \neq 1$.

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Here is a variation of Jonas's proof.

Let $\mu$ be the product of the $\mu_{T_i}$, which I will denote $\mu_i$, and let $K$ be the ground field.

Since $\mu(T)=0$, the ring $K[T]$ is a quotient of $A:=K[X]/(\mu)$, and $\mathbb{V}$ is an $A$-module.

By the Chinese Remainder Theorem, $A$ is isomorphic to the product of the $A_i:=K[X]/(\mu_i)$.

The element $e_i$ of $A$ whose $j$-th component is $\delta_{ij}$ acts by the identity on $W_i$, and by $0$ on $W_j$ for $j\neq i$.

For $i=1,\dots,k$, let $w_i$ be in $W_i$, and assume $\sum w_i=0$.

By applying $e_j$ to the above equality, we get $w_j=0$.

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