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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?

t.b.
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user7980
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    If $V$ is a $T$-invariant subspace, then the minimal polynomial of the endomorphism of $V$ induced by $T$ divides the minimal polynomial of $T$. [By the way, there is a typo in the title.] – Pierre-Yves Gaillard Aug 10 '11 at 08:33
  • Thank you. So the existence of a non-trivial $T$ invariant subspace is equivalent to the existence of a root for the characteristic polynomial existing in the base field? – user7980 Aug 10 '11 at 08:56
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    You’re welcome. The existence of a non-trivial T invariant subspace is equivalent to the existence of a non-trivial factor of the characteristic polynomial existing in the base field. [I think I can make this (slightly) more precise if you want.] – Pierre-Yves Gaillard Aug 10 '11 at 09:02
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    Let $f$ be in $K[X]$, where $K$ is a field and $X$ an indeterminate. The ideals of the ring $K[X]/(f)$ correspond to the ideals of $K[X]$ which contain $f$, and thus to the factors of $f$. – Pierre-Yves Gaillard Aug 10 '11 at 09:16
  • @Pierre: what does "non-trivial factor" mean? In fact, pick the identity $\mathrm{id} \colon \mathbb{K}^2 \to \mathbb{K}^2$: its minimal polynomial is irreducible, but every line is a non-trivial invariant subspace. – Andrea Aug 10 '11 at 09:38
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    @Andrea: We're talking about companion matrices. (The minimal and characteristic polynomials must coincide.) (Look at the beginning of the question.) – Pierre-Yves Gaillard Aug 10 '11 at 09:47
  • Certainly. I thought your observation was general and not only for companion matrices. – Andrea Aug 10 '11 at 09:57
  • Fixed the typo in the title. – Gerry Myerson Aug 10 '11 at 13:05
  • 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. – Marc van Leeuwen Jan 12 '15 at 13:47

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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 $x$.

Pierre-Yves Gaillard
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