21
$\begingroup$

Let $R$ be an integral domain and $F$ a finitely generated free module over $R$. For a linear transformation $\alpha\in\operatorname{End}_R(F)$, the characteristic polynomial is \begin{equation} p_\alpha(t)=\det(t-\alpha)\in R[t]. \end{equation}

Similar transformations have same characteristic polynomial. However, these polynomials fail to characterize similarity, in the sense that non-similar transformations can have the same characteristic polynomial.

So my question is: What can we say about two transformations $\alpha$ and $\beta$ that share the same polynomial?

Obviously, $\alpha$ and $\beta$ have the same spectrum and same algebraic multiplicity for each eigenvalue.

But unless $R$ is an algebraically closed field--in this case the polynomial is determined by the roots and their multiplicities--we should be able to say more about $\alpha$ and $\beta$. Also we should know more than traces and determinants since they are just two of the coefficients.

Can someone give a hint? Thanks!

$\endgroup$
4
  • $\begingroup$ +1 This is one of those questions I always wanted to get around to sorting out in my head. I'm pretty sure the answer is here somewhere in this elementary construction: given a transformation $T$ of a module $M$, you can look upon $M$ as a $K[T]$ module, where $T^n\cdot m:=T^n(m)$ and everything else is defined by linearity. The characteristic and minimal polynomials have to do with the kernel of $K[x]\rightarrow K[T]$, since $K[x]$ is a PID. I'm not sure why the two polynomials carry different information, and I'm also not sure of what happens if you replace $K$ with an integral domain. $\endgroup$
    – rschwieb
    Feb 19 '13 at 15:25
  • 1
    $\begingroup$ You also might want to try to solve the question for $R$ a field before you try for $R$ a domain. I know from your past questions that domains seem to be the star of the show for you, but seriously, fields are "the Cadillac of rings"! :) $\endgroup$
    – rschwieb
    Feb 19 '13 at 15:29
  • $\begingroup$ @rschwieb: If fields are the Cadillac, what are algebraically closed fields? $\endgroup$ Feb 19 '13 at 16:10
  • 1
    $\begingroup$ @MarcvanLeeuwen Even Cadillacs have special features! Leather interior maybe? $\endgroup$
    – rschwieb
    Feb 19 '13 at 17:26
5
$\begingroup$

The degree to which the characteristic polynomial fails to charcterise similarity can be illustrated, for the case where $R$ is a field, by the rational canonical form, which in this case does characterise similarity. The rational canonical form is determined by a uniquely defined sequence of monic polynomials (invariant factors), each one a multiple of any one of polynomials that follow (I prefer this to the more common requirement of dividing what follows); the sequence can have any length, but it can be thought of ending with an indefinite repetition of the constant polynomial $1$ (just like partitions of an integer are weakly decreasing sequences of integers that can be thought of as ending with endless occurrences of $0$). The first invariant factor (in the order I chose) is the minimal polynomial, and the product of all invariant factors is the characteristic polynomial.

Now if you fix the characteristic polynomial, its irreducible factors in $R[X]$ are determined; all of them must occur as a factor of the minimal polynomial. Indeed the multiplicity of a fixed irreducible polynomial $P$ in the invariant factors must be weakly decreasing, so these multiplicities form a partition of the multiplicity of $P$ in the characteristic polynomial. Choosing a partition for each occurring irreducible factor $P$ is exactly the freedom one has for choosing a rational canonical form, and therefore determines the number of similarity classes of transformations with a given characteristic polynomial.

The "regular" case is where the minimal polynomial is equal to the characteristic polynomial; this is forced when the characteristic polynomial is squarefree. In this case the centraliser of $\alpha$ (in the endomorpism algebra) is equal to the set of polynomials in $\alpha$, which is of dimension $n$ (the same as that of the vector space on which $\alpha$ acts). I think that in all other cases the dimension of the centraliser is strictly larger than $n$, whence the similarity class is smaller than in the regular case; this intuitively explains why there aren't any polynomial functions that can separate the "singular" cases from the regular case with the same characteristic polynomial.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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