# Irreducible characteristic polynomial of a linear transformation

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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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
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
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
@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

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 $x$.

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Claim. If $V$ is a finite-dimensional $k$-vector space and $A:V\to V$ is a linear transformation, then the characteristic polynomial $P_A$ is irreducible if and only if $A$ has no proper nontrivial $A$-invariant subspaces.

Proof. One direction is trivial: if there is an invariant subspace, the restriction of $A$ to that subspace has characteristic polynomial dividing $P_A$.

In the other direction, suppose $P_A$ factors as $QR$. Note that Cayley-Hamilton tells us $P_A(A)=0$, so $Q(A)R(A)=0$.

Now assume $A$ has no proper nontrivial $A$-invariant subspaces and consider the image of any polynomial $f(A)$. $A$ commutes with each $A^n$ and thus with $f(A)$, so $Af(A)x=f(A)Ax$, and the image of $f(A)$ is an $A$-invariant subspace, and thus is $0$ or $V$, i.e. $f(A)$ is either invertible or 0.

Now if $Q(A)R(A)=0$ they can't both be invertible, so at least one (WLOG $Q(A)$) is zero. This is a monic polynomial of strictly lower degree than $\dim V$.

Set $n=\deg V$, fix some nonzero $x$, and consider the subspace $W=\langle x,Ax,A^2x,\dots,A^{n-1}x\rangle$. $W$ is proper (it contains $x$) and nontrivial (it is generated by less than $\dim V$ vectors). Since $A^nx$ is a linear combination of lower powers, $A^nx\in W$. Thus $A$ maps each of these generators into $W$, so $W$ is a proper nontrivial $A$-invariant subspace.

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This is nice, although it spends most time doing the direction that the question is not about ($P_A$ reducible, then there exists a nontrivial invariant subspace; by the way "not irreducible" is not quite good enough, for $\dim=0$). I would do that direction somewhat differently: if $P_A$ is reducible then it has a proper irreducible factor $f$; as $f$ divides the minimal polynomial, $f[A]$ is not invertible; any nonzero $v\in\ker(f[A])$ spans an invariant subspace of dimension $\deg(f)<\deg(P_A)=\dim(V)$ (as in your final paragraph), contradiction. – Marc van Leeuwen Jan 12 '15 at 13:56