I'm reading Hoffman's "Linear Algebra" and this question comes to my mind: suppose $V$ is a finite-dimensional vector space on a field $F$, and $T$ a linear operator on $V$, when does a matrix has Cyclic Decomposition, and when does it has a Rational Form?

The **Cyclic Decomposition** Theorem (Hoffman, Thm 7.3) says:

> 
> Let $T$ be a linear operator on a finite-dimensional vector space $V$
>  and let $W\_0$ be a proper $T$-admissible subspace of $V$. There exist
>  non-zero vectors $a\_1, \dots , a\_r$ in $V$ with respective
>  $T$-annihilators $p\_1, \dots , p\_r$ such that $V = W\_0 \oplus
>  Z(\alpha\_1; T) \oplus \dots \oplus Z (\alpha\_r ; T)$ ;
> 
> 
> 

So the Cyclic Decomposition exists iff there is a $T$-admissible proper subspace $W\_0 \subsetneq V$ to start from. Then what is the the condition for $W\_0$'s existence?

A **Rational Form** is actually the Cyclic Decomposition with $W\_0=\{0\}$. So does this mean the condition for the existence of a rational form is that $T$'s character values are all in field $F$? 

For example a $n\times n$ matrix $T$ on real number field $\mathbb R$ has a rational form, so long as all the characteristic values of $T$ are all real numbers?