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I am trying to understand why the linear transformation $T$ corresponding to the companion matrix of the minimal polynomial of an irreducible polynomial over $\mathbb{Q}[x]$ (for instance $x^3-x-1$) has no non-trivial $T$-invariant subspaces.

I know the minimal polynomial of $T$ is equal to the characteristic polynomial of $T$ and furthermore $T:\mathbb{Q}^4\rightarrow \mathbb{Q}^4$ has a cyclic vector $\alpha$ ( that is $\exists \alpha \in \mathbb{Q}^4$ such that $\mathbb{Q}^4 = \{ g(T)\alpha : g \in \mathbb{Q}[x]\}$.

Question: I have not been able to find a specific theorem in my textbook that tells you that when the minimal polynomial is irreducible then it has no non trivial T-invariant subspaces. My question is does there exist such a result or can we make a similar statement when $\mathbb{Q}$ is replaced by any field that is not algebraically closed?

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    The reason you could not find a statement saying that when the minimal polynomial is irreducible then it has no nontrivial $T$-invariant subspaces, is that this is obviously false. Consider a scalar multiple of the identity, whose minmal polynomial has degree $1$ whence is irreducible.2015-01-12

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Let $K$ be a field, $X$ an indeterminate, $f\in K[X]$ a degree $n$ monic polynomial, and $A$ the companion matrix. Then there are natural bijections between

  • the $A$-invariant subspaces of $K^n$,

  • the ideals of $R:=K[X]/(f)$,

  • the ideals of $K[X]$ which contain $f$,

  • the monic divisors of $f$.

To see this, let $x$ be image of $x$ in $R$, and observe that $A$ is the matrix of the multiplication by $x$ in the basis formed by the degree $ < n$ monomials in $x$.