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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.

Alphonse
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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 $r=1$).

Marc van Leeuwen
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  • 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
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    I've added an sentence explaining my point. – Marc van Leeuwen Mar 26 '17 at 06:10
  • Thank you very much ! I see now that $W$ is only a subspace of $ker(P_1(T))$. I've understood everything, except one thing: why is $P_1$ the minimal degree monic polynomial such that $P_1T=0$ ? I guess it is by irreducibility of $P_1$, but I'm not sure. I tried to do an Euclidean division of $P_1$ by some smaller degree polynomial $Q$ such that $Q(T)v = 0$, i.e. $P_1 = QR_1+S_1$, but I don't see why $S_1=0$ (which would lead to the conclusion). Thank you very much for your help! – Alphonse Mar 26 '17 at 08:52
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    Yes, what you must use is the standard argument that the minimal degree (monic) polynomial $Q$ with $QT=0$ 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 $ST=0$ (since $(QR_1)T =R_1T =R_1T=0$), 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 09:53
  • Thank you very much for your detailed comments. I appreciate your help. – Alphonse Mar 26 '17 at 10:16