# Minimal and characteristic polynomial have same set of irreducible factors

Let $V$ be a vector space over a field $F$ (not necessarily algebraically closed), and $T:V\rightarrow W$ be a $F$-linear map. If $c(T)$ and $m(T)$ denote the characteristic and minimal polynomial respectively, then I want to show that $c(T)$ and $m(T)$ have the same set of irreducible factors.

I know that the claim is true if $F$ is algebraically closed, because then the two polynomials have the same roots. The typical reasoning one gives when $F$ is not algebraically closed is: Take an algebraic closure of $F$, say $K$, and look at $m(T)$ and $c(T)$ in $K$ to conclude the proof. But as far as I know, the proof that any root of $c(T)$ is a root of $m(T)$ requires the fact that $V$ is a $K$-vector space. But we only know that $V$ is an $F$-vector space. I can't understand how to proceed after this.

• You must assume $V=W$ ! – Georges Elencwajg Sep 6 '13 at 8:27

It is easy to see that any root in the (non-algebraically closed) field $F$ of the characteristic polynomial is also a root of the minimal polynomial; algebraic closure plays no role here. Such a root$~\lambda$ is a (true) eigenvalue with an actual eigenvector$~v$ in the original vector space: $T\cdot v=\lambda v$. But then $P[T]\cdot v=P[\lambda]v$ for any polynomial $P[T]$ of$~T$, an in particular $0=0\cdot v= m[T]\cdot v=m[\lambda]v$, and since $v\neq0$ it must be that $m[\lambda]=0$ so $\lambda$ is a root of the minimal polynomial $~m$.
However this does not show that other irreducible factors $p$ of the characteristic polynomial divide the minimal polynomial. For that the easiest argument is to extend the field to $K$ so that $p$ admits a root$~\lambda$ in$~K$ (there is no need for a further extension); choosing a matrix for $T$ it defines an endomorphism of $K^n$ with characteristic polynomial $c$ and (somewhat less obviously) minimal polynomial$~m$. Then by the above $\lambda\in K$ is a root of$~m$, which therefore over$~F$ must have a factor$~p$ (as this is the minimal polynomial of$~\lambda$ in the sense of field theory). However, one can give a proof avoiding field extensions, though it is a bit more work. I gave one in (the last part of) this answer.
In basis-free jargon, what you do is consider the $K$-linear map induced by $T$ on the tensor product $V \otimes K$. If you don't know about tensor products yet, this amounts to using a basis for $V$ to represent $T$ as an $n \times n$-matrix $M$ (where $n$ is the dimension of $V$) and then consider the $K$-linear map on $K^n$ represented by $M$.