Q:

proof about commutative operators and T-cyclic vectors

Let $V$ be a finite dimensional vector space over $F$. Let $T:V \to V$ be a linear operator.
Prove that if every linear operator $U$ which commutes with $T$ is a polynomial of $T$, than $T$ has a $T$-cyclic vector.
I don't really know where to start...
can someone please point me in the right direction?

A:

With some general theory this can be done quite painlessly. By the structure theorem for finitely generated modules over a principal ideal domain (with $F[X]$ as PID, and $V$ considered as $F[X]$-module with $X$ acting as $T$), $V$ decomposes as a finite direct sum of cyclic modules; the decomposition is not unique, but the number and types of the modules are unique (and so $V$ has a cyclic vector if and only if there is (at most) one cyclic module in such a decomposition). The cyclic factors are either isomorphic to $F[X]$ as module over itself (but this cannot happen for $V$ because it is finite dimensional over $F$), or of the form $F[X]/P$ for a non-constant monic polynomial $P\in F[X]$; moreover these polynomials $P_i$ in the successive cyclic modules (called the invariant factors of the $F[X]$ module $V$) divide each other: $P_i\mid P_{i+1}$ whenever both polynomials exist.
Now the question is to show that if all linear operators commuting with $T$ are polynomials in $T$ (in other words they are just scalar multiplications in the $F[X]$-module, where "scalar" means element of $F[X]$) then there cannot be two or more cyclic modules in the decomposition (the converse is also true, see this question). So assume a decomposition given by the theorem has at least two factors $F[X]/P_1$, $F[X]/P_2$, with $1\neq P_1\mid P_2$. Since the factors in the decomposition are submodules, every scalar multiplication stabilises them; it will suffice to find a $F[X]$-module morphism $V\to V$ that does not stabilise all factors (the module morphism property says that it is $F$-linear and commutes with $T$). But we have a non-zero morphism $\pi:F[X]/P_2\to F[X]/P_1$ of reducing modulo $P_1$, which is defined because $P_2$ is a multiple of $P_1$. To make this into a $F[X]$-module endomorphism $V$, it suffices to first project $V\to F[X]/P_2$ parallel to all other cyclic factors, then apply$~\pi$, and finally inject the factor $F[X]/P_1$ back into $V$. The resulting endomorphism is easily seen to not stabilise the cyclic factor $F[X]/P_2$, as desired. QED

A:

To my understanding this is the so called cyclic vector theorem. There are many versions of the proof accessible. One quite comprehensive comes here:
http://planetmath.org/proofofcyclicvectortheorem
Would hardly make sense to copy down all to here.
Hope this helps.

