When matrices commute

There is a little room for repeat eigenvalues, as long as we have nontrivial Jordan blocks. For the following, if an integral square matrix commutes with $A_j,$ it is a (rational) polynomial in $A_j$: $$ A_2 \; = \; \left( \begin{array}{rr} 1 & 1 \\\ 0 & 1 \end{array} \right) , $$

$$ A_3 \; = \; \left( \begin{array}{rrr} 1 & 1 & 0 \\\ 0 & 1 & 1 \\\ 0 & 0 & 1 \end{array} \right) , $$

$$ A_4 \; = \; \left( \begin{array}{cccc} 0 & -1 & 1 & 0 \\\ 1 & 0 & 0 & 1 \\\ 0 & 0 & 0 & -1 \\\ 0 & 0 & 1 & 0 \end{array} \right). $$

EDIT : it seems reasonable to conjecture that the full set of $A \in SL_n(\mathbb Z)$ for which the statement is true is $ A \in SL_n(\mathbb Z)$ such that, should there be any eigenvalue(s) of multiplicity larger than one, all occurrences of that eigenvalue must fit into a single Jordan block. Richard would know.

EDIT, 20 November 2011: the conjecture above is true, and does not use integers, it is just about matrices over the complex numbers. This is Corollary 1 to Theorem 2, on page 222 of The theory of matrices, Volume 1 by Feliks Ruvimovich Gantmakher. It reads:

Corollary 1 to Theorem 2: All the matrices that are permutable with $A$ can be expressed as polynomials in $A$ if and only if $n_1=n,$ i.e. if all the elementary divisors of $A$ are coprime in pairs.

SO, the following two conditions, for a square matrix $M$ with real or complex entries, are equivalent:

(I) All matrices that commute with $M$ can be written as a polynomial in $M.$

(II) The characteristic polynomial and the minimal polynomial of $M$ are the same.

(1) Let $A$ be a commutative ring, and let $M$ and $N$ be $A$-modules. If the natural morphism from $A$ to $\text{End}_A(M\oplus N)$ is surjective, then the annihilators of $M$ and $N$ are comaximal.

Indeed, this comaximality is the condition for the projectors attached the given direct sum decomposition to be in the image.

Assume now that $A$ is a principal ideal domain. Let $(G,+)$ be the Grothendieck group of the category $C$ of finitely generated torsion $A$-modules, and let $(H,\cdot)$ be the group of (nonzero) fractional ideals of $A$.

There is a (clearly unique) morphism from $G$ to $H$ which maps $A/\mathfrak a$ to $\mathfrak a$.

It is easy to see that the fractional ideal attached to $M\in C$ is in fact integral. Call it the characteristic ideal of $M$. Moreover we have in view of (1):

(2) The natural morphism from $A$ to $\text{End}_A(M)$ is surjective, if and only if the characteristic ideal of $M$ coincides with the annihilator of $M$.

Assume now that $A=K[X]$, where $K$ is a field and $X$ an indeterminate, and that $V$ is a finite dimensional $K$-vector space equipped with an endomorphism $a$. Then the characteristic ideal of $V$ is generated by the characteristic polynomial of $a$, and (1) and (2) imply:

The characteristic polynomial of $a$ coincides with its minimal polynomial if and only if any endomorphism of $V$ commuting with $a$ is in $K[a]$.