# 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.

• – Watson Nov 26 '18 at 12:30

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 $r=1$).
• Thank you. Don't you mean $ker(P_1(T))$ at the beginning instead of $ker(P(T))$ ? For me, $P(T)=0$. – Alphonse Mar 25 '17 at 11:14
• Right, I meant that. (The name $P$ did not make me think of a minimal polynomial, a weak excuse.) – Marc van Leeuwen Mar 25 '17 at 13:04
• Thank you (I think that some $P$'s should still be $P_1$'s). But how do you show that the $T^i (v)$ are a basis for $W := ker(P_1(T))$ ? I tried to show that they are linearly independent, without success. I'm not sure how to compute the dimension of $W$, by the way. Thank you :-) – Alphonse Mar 25 '17 at 15:02
• Yes, what you must use is the standard argument that the minimal degree (monic) polynomial $Q$ with $Q[T](v)=0$ divides any other polynomial with this property (since $P_1$ has the property and is irreducible, this will then give $Q=P_1$). Do the Euclidean division that you indicated. Now $S=P_1-QR_1$ also has $S[T](v)=0$ (since $(QR_1)[T](v) =R_1[T](Q[T](v)) =R_1[T](0)=0$), but by minimality of $Q$ this can only mean that $S=0$. It is essentially the same argument that shows for instance that the least common multiple of polynomials $A,B$ divides any other common multiple of $A,B$. – Marc van Leeuwen Mar 26 '17 at 9:53