# Is it always possible to “separate” the eigenvalues of an integer matrix?

Call a square matrix Galois-irreducible if all its eigenvalues are Galois conjugates of each other.

Let $M$ be an integer $n\times n$ matrix which is not Galois-irreducible. Is it always possible to find an integer matrix $S$ such that $S^{-1}MS=diag(A_1,...,A_k)$ is a block diagonal matrix with Galois-irreducible matrices $A_i$ ?

Intuitively this should be true, but I have no idea how to construct such a matrix $S$.

Note that this is not true when replacing everywhere "Galois-irreducibility" by "irreducibility of the characteristic polynomial" because any non-trivial Jordan blocks like $\begin{pmatrix}\lambda&1\\ 0&\lambda\\ \end{pmatrix}$ for, say, $\lambda\in\mathbb Z$, would yield a counterexample.

In a similar vein:

Suppose $M$ is this time Galois-irreducible, but with each eigenvalue of multiplicity $k$ and trivial Jordan blocks. Does there exist $S$ as above, maybe even such that $A_1=\cdots=A_k$?

(It is clear that for non-trivial Jordan blocks the latter isn't possible.)

• Do you also want the $A_i$ to have integer entries? And $S^{-1}$? – Qiaochu Yuan Dec 13 '15 at 22:44
• No I don't have the illusion that there might be such an $S$ with determinant $1$. So, well, I could have said as well "rational" instead of "integer". BTW, as I write this comment, I just saw Eric's comments, so... – Wolfgang Dec 14 '15 at 7:08
• Okay, if you're happy with rational everywhere then you can just appeal to the structure theorem for f.g. modules over a PID (I edited with this observation). – Qiaochu Yuan Dec 14 '15 at 7:47
• Well, thinking again: there is thus no use of requiring $S$ integer. But once rational $A_i$'s are found (in theory), we can obviously find for each $i$ another base in which $A_i$ has integer entries. – Wolfgang Dec 14 '15 at 11:42

Think of $M$ first as a linear operator acting on $V = \mathbb{Q}^n$. Pass to a splitting field $K$ and consider the induced action on $V \otimes K$. This splits up into a direct sum of generalized eigenspaces of $M$, which are also then permuted by the Galois action of $G = \text{Gal}(K/\mathbb{Q})$ into orbits. By Galois descent we can find $G$-invariant bases of these sums over each Galois orbit of generalized eigenspaces. With respect to these bases, $M$ acts on each orbit by a Galois-irreducible rational matrix. If $M$ has no nontrivial Jordan blocks then you can replace "generalized eigenspaces" with "eigenspaces" and I think with a little more fiddling you get your second statement as well.
So that's over $\mathbb{Q}$. I'm less sure what to do over $\mathbb{Z}$. Presumably you should look at the induced action on $\mathbb{Z}^n \otimes \mathcal{O}_K$.
Edit: Again over $\mathbb{Q}$, you can avoid appealing to Galois descent as follows: the structure theorem for f.g. modules over a PID implies that $V$ above, as a $\mathbb{Q}[x]$-module (where $x$ acts by $M$), breaks up as a direct sum of modules of the form $\mathbb{Q}[x]/f(x)^m$ where $f$ is irreducible. Multiplication by $x$ always acts by a Galois-irreducible matrix on such submodules. Trivial Jordan blocks means we can take $m = 1$.
• Aah, actually, there are two interpretations of the question. If you want $S$ to have integer entries but don't care about whether its inverse does, then just take Quaicho's $S$ and scale it. If on the other hand you want the inverse to have integer entries too then doesn't a matrix like $(3,1;0,1)$ kill this question immediately? Not diagonalisable mod 2 so not diagonalisable over Z in this strict sense. – eric Dec 13 '15 at 22:07
If a matrix $M$ is not Galois-irreducile, then its minimal annihilating polynomial $\mu_M(x)$ factors as a product of two coprime non-constant polynomials, $\mu_M(x)=p_1(x)p_2(x)$. Then $\mathbb Q^n=\mathop{\rm Ker} p_1(M)\oplus \mathop{\rm Ker} p_2(M)$, both subspaces being $M$-invariant, which leads to the required representation.