Q:

Irreducible factor of the minimal polynomial

Let $F$ be a field of characteristic $0$, $V = F^n$ for some $n$, and $T : V \to V$ a linear transformation. If the minimal polynomial $P(x) \in F[x]$ of $T$ has an irreducible factor $P_1(x)$ of degree $r$, is it true that $V$ has a $T$-invariant subspace of dimension $r$?
I tried to write $P(x) = \prod_i P_i^{e_i}(x)$ as product of irreducible polynomials, and then $V = \bigoplus_i \ker(P_i^{e_i}(A))$. I'm not sure how to continue.

A:

Yes: $\ker(P_1[T])$ is a nonzero subspace (or else $P_1$ would not divide the minimal polynomial$~P$, more precisely $P/P_1$ would still be an annihilating polynomial for$~T$, which it is not), and for any nonzero vector $v\in\ker(P_1[T])$ one has that $P_1$ is the minimal degree monic polynomial such that $P_1[T](v)=0$. But then the vectors $T^i(v)$ for $0\leq i<r=\deg(P_1)$ form a basis of a $T$-invariant subspace$~W$ (of $V$ and indeed of $\ker(P_1[T])$), which then of course has dimension$~r$. This is so because having a linear dependence relation $c_0v+c_1T(v)+\cdots+c_dT^d(v)=0$ precisely means that $Q[T](v)=0$ for the polynomial $Q=c_0+c_1X+\cdots+c_dX^d$, which does not happen (unless $Q=0$) for $d<r$, but does happen (with $c_d=1$) for $d=r$; the latter implies $T^r(v)\in W$ and thereby $T(W)\subseteq W$.
This is a straightforward generalisation of the fact (for $r=1$) that every root of the minimal polynomial is an eigenvalue (any corresponding eigenvector spans a $T$-invariant subspace of dimension $r=1$).

