Every $v \in V - \{ 0 \}$ is cyclic iff the characteristic polynomial of $T : V \to V$ is irreducible over $F$ Let $V\neq \{0\}$ be a vector space over $F$, and $T$ a linear operator on $V$. Prove that every $0\neq v \in V$ is a cyclic vector if and only if the characteristic polynomial of $T$ is irreducible over $F$.
I can't seem to get anywhere on either side... Can someone please help?
 A: One direction is easy: if $v\neq0$ is not a cyclic vector, then the span of the vectors $T^i(v)$ for $i\geq0$ is not the whole space but is $T$-stable; therefore the characteristic polynomial of $T$ restricted to that space is a nontrivial strict divisor of the characteristic polynomial of$~T$, which is therefore reducible.
The other direction is harder; it requires showing that if the characteristic polynomial is reducible then there exists a nonzero non-cyclic vector for $T$. I seem to need the following trick: pick any nonzero vector; if it is non-cyclic, we are done, so assuming it is cyclic the minimal polynomial in $T$ that kills this vector has degree $\dim V$, so it must be equal to the characteristic polynomial. Then the minimal polynomial is equal to the characteristic polynomial, and hence (also) reducible. Now take a proper nontrivial factor $P$ of the minimal polynomial and evaluate it in $T$ giving a linear operator $P(T)$. Using minimality of the minimal polynomial show that $P(T)$ is neither zero nor invertible; its kernel then is a nonzero proper $T$-stable subspace, and the nonzero vectors in it are not cyclic.
Added. Thinking about this, it strikes me that the statement to prove, although pleasingly compact, is not really the best formulation; the situation is more completely described as follows, with the statement as corollary.
Proposition. Let the operator $T$ on the finite dimensional $F$-vector space $V$ have minimal polynomial $\mu_T\in F[X]$. If $\deg\mu_T<\dim V$ then $V$ has no cyclic vectors at all, if $\mu_T$ is reducible then $V$ has nonzero non-cyclic vectors, and when neither of these is the case one has $\deg\mu_T=\dim V$ and all nonzero vectors of$~V$ are cyclic.
Then by the Cayley-Hamilton theorem (which is not used for the proposition itself), if the characteristic polynomial irreducible then the minimal polynomial must be equal to it and we are in the final case; conversely if all nonzero vectors are cyclic then $\mu_T$ must be irreducible (or $1$, if $\dim V=0$) and equal to the characteristic polynomial.
Proof. For a vector$~v$ denote by $\mu_{T,v}$ the monic polynomial$~P$ in$~F[X]$ of least degree for which $P(T)(v)=0$; then $\deg\mu_{T,v}\leq\dim V$, and $v$ is a cyclic vector if and only if $\deg\mu_{T,v}=\dim V$. Since $\mu_T(T)(v)=0$ one always has $\mu_{T,v}\mid\mu_T$, which implies that if $\deg\mu_T<\dim V$ there cannot be any cyclic vectors. If $\mu_T=PQ$ with $P,Q\in F[X]$ non-constant, then neither $P(T)$ nor $Q(T)$ are zero operators by minimality of $\mu_T$; but their composition is zero, so the subspace $\ker P(T)$ contains the image of $Q(T)$ and is a nonzero proper $T$-stable subspace, whose nonzero vectors are therefore non-cyclic; this establishes the second case. Conversely if $\mu_T$ is irreducible then every nonzero vector$~v$ has $\mu_{T,v}=\mu_T$, so in particular $\deg\mu_T\leq\dim V$, and if $\deg\mu_T=\dim V$ these vectors are all cyclic.
Note that the proposition does not say whether there are any cyclic vectors when $\mu_T$ has degree$~\dim V$ but is reducible; this is in fact true, but requires more detailed considerations to prove.
