Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
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 8:33
    
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 8:56
1  
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 9:02
1  
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 9:16
1  
@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 9:47

1 Answer 1

up vote 2 down vote accepted

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 $x$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.