# A criteria for Jordan decomposition of a matrix over a general ring

When I look at different proofs of the Jordan-Chevalley decomposition of a matrix, the minimal hypothesis I usually found is about the perfection of the field over which such decomposition occurs (e.g. Wikipedia article). But it seems to me that the gist of the proofs is not about the field but about the type of characteristic polynomial of the matrix.

More precisely, by using the two following definitions for a general ring $$A$$

Definition 1 A $$n\times n$$-matrix $$M$$ with coefficients in the ring A is semisimple if $$A^n$$ is a semi-simple $$A[M]$$-module (here I am using the definition provided in the answer of @ಠ_ಠ below)

Definition 2 A separable polynomial $$P$$ of $$A[X]$$ is a polynomial whose discriminant is invertible in $$A$$

Then is the following statement true?

A matrix $$M$$ over $$A$$ have a unique Jordan-Chevalley decomposition if its characteristic polynomial divides a power of a separable polynomial.

Edit By Jordan decomposition I mean here the additive one, i.e. there exist $$M_s$$ and $$M_n$$, two $$n\times n$$ $$A$$-matrices respectively semi-simple and nilpotent, such that $$M=M_s+M_n$$

If it is not, what kind of general property should the ring $$A$$ have to make it true (like $$A$$ is a domain, etc.)

Edit 2 I believe my definition 1 is equivalent in a domain to the fact that the matrix is diagonalizable in an algebraic extension of the fraction field which is equivalent to the fact that the matrix is cancelled by a separable polynomial (definition 2) in $$A[X]$$

• @GetOffTheInternet Done. (Ps: I love your pseudo) – brunoh Oct 14 '18 at 13:37

Let $$A\in M_n(R)$$ where $$R$$ is some commutative ring, $$p$$ be the "square free" polynomial that divides $$\chi_A$$ (the characteristic polynomial of $$A$$) and $$p'$$ be the derivative of $$p$$.

We consider the pseudo Newton algorithm

$$X_0=A , X_{k+1}=X_k-p(X_k)(p'(X_k))^{-1}$$. When $$R$$ is a perfect field, after a finite number of iterations, $$X_k=D$$, the semi simple part of $$A$$.

Of course, we have to give conditions on $$R$$, for example, as assumed by Minseon Shin,

"Let $$R$$ be an integral domain whose fraction field $$K$$ is perfect".

We can chain the iterations iff $$p'(X_k)$$ is invertible over $$M_n(R)$$, that is, if $$\det(p'(X_k))$$ is an invertible of $$R$$. In fact, at each iteration, $$\det(p'(X_k))=\det(p'(A))$$ and the invertibility of $$\det(p'(A))$$ suffices. It is not difficult to see that $$\det(p'(A))$$ is a divisor of a power of $$discrim(p)$$.

Finally, if $$discrim(p)$$ is an invertible (that is equivalent to the condition considered by the OP), then we can chain the iterations.

When $$R=\mathbb{Z}/m\mathbb{Z}$$, the entire algorithm seems to work; I did not find any counter-examples.

Assume that $$R$$ is an integral domain. If $$p\in R[x]$$, then $$discrim(p)\not= 0$$ implies that $$p,p'$$ have no common roots over an algebraically closed field containing $$R$$, that implies that $$p$$ has only simple roots. If moreover, $$discrim(p)$$ is invertible and $$p$$ divides $$\chi_A$$, then I think that the above algorithm formally works and that we obtain $$D$$ s.t. $$p(D)=0$$ and $$A-D$$ nilpotent.

However, practically, how to find such a $$p$$ ? A priori, $$R[x]$$ is not an UFD; on the other hand,it seems difficult to work over a field that contains $$R$$ and the roots of $$\chi_A$$; how to prove that $$p$$ divides $$\chi_A$$ if $$R$$ is not euclidean or not pincipal?

Anyway, in general, $$p=\dfrac{\chi_A}{gcd(\chi_A,\chi'_A)}$$. Note that if $$R$$ is an UFD, then the gcd always exists in $$R[x]$$; yet, in general, there is no algorithm to do that.

Finally, if $$R$$ is an Euclidean domain, I think that we can explicitly calculate $$D,N$$. However, in theory, perhaps the hypothesis UFD suffices...

• Thank you for your effort +1. The gist of the question is about trying to find the most general condition on the ring considering the specific matrix we want to decompose. I know from Bourbaki that for the decomposition to work for any matrix in the ring it is necessary to have a perfect field of fraction. But looking at the proofs (especially the one you gave – brunoh Oct 16 '18 at 21:13
• (Continuation of my previous comment) I got the impression that my condition was sufficient for a matrix in any domain. Indeed I think considering that my first definition is equivalent to the fact that a semi-simple matrix is cancelled by a separable polynomial (definition 2), your proof seems to work fine even if the ring is just supposed to be a domain. Am i right here ? – brunoh Oct 16 '18 at 21:20
• Your Edit is clear. I was not looking for a constructive criteria but one that allows me to understand what is truly at stake in the different proofs of the decomposition, without any unnecessary hypothesis on the ambient ring. I will wait a bit in case a more definite answer comes but it seems you confirm my criteria is not wrong. – brunoh Oct 18 '18 at 16:08
• @brunoh , thanks for the bonus. – loup blanc Oct 19 '18 at 20:26

If $K$ is a commutative ring and $V$ is a $K$-module, then a choice of endomorphism $f \in \text{End}_K(V)$ is the same as a representation of the $K$-algebra $K[f]$ on $V$. This is proved in Aluffi's Algebra text in section VI.7.1. Then we say $f$ is simple (resp. semisimple) if $V$ is simple (resp. semisimple) as a $K[f]$-module.

Sorry I can't help with the rest of your questions.

(This is only a partial answer and I suspect it only confirms what you already knew but it was an interesting exercise for me and I thought I'd write it up.)

I will use the following interpretation of your question:

Let $$A$$ be an integral domain whose fraction field $$K$$ is perfect. For a matrix $$M \in \mathrm{Mat}_{n \times n}(A)$$, let $$M = M_{\mathrm{ss}} + M_{\mathrm{n}}$$ be its Jordan decomposition over $$K$$ (so $$M_{\mathrm{ss}},M_{\mathrm{n}} \in \mathrm{Mat}_{n \times n}(K)$$). We may ask whether $$M_{\mathrm{ss}},M_{\mathrm{n}} \in \mathrm{Mat}_{n \times n}(A)$$.

Suppose $$A$$ is normal. Let $$p_{M} \in A[t]$$ be the characteristic polynomial of $$M$$. Let $$\overline{K}$$ be an algebraic closure of $$K$$, let $$\overline{A}$$ be the integral closure of $$A$$ in $$\overline{K}$$, and let $$\lambda_{1},\dotsc,\lambda_{r} \in \overline{K}$$ be the distinct roots of $$p_{M}$$. Since $$\overline{A}$$ is normal, we in fact have $$\lambda_{i} \in \overline{A}$$. If every pairwise difference $$\lambda_{i} - \lambda_{j}$$ (with $$i \ne j$$) is invertible in $$\overline{A}$$, then $$M_{\mathrm{ss}},M_{\mathrm{n}} \in \mathrm{Mat}_{n \times n}(A)$$.

Remark: Your condition that $$p_{M}$$ divides a power of a polynomial with invertible discriminant is nicer because it makes sense over any ring, but if $$A$$ is a normal domain, then I think it implies my condition that $$\lambda_{i} - \lambda_{j}$$ is invertible. It's not clear to me whether the converse is true if $$A$$ is a domain which is not a UFD.

Proof: Let $$K'/K$$ be a finite Galois extension over which $$p_{M}$$ splits, and let $$A'$$ be the integral closure of $$K$$ in $$K'$$. Since $$A$$ is normal, the inclusion $$A \subseteq K \cap A'$$ is an equality. Thus we may reduce to the case when $$p_{M}$$ splits into linear factors over $$A$$, say $$p_{M}(t) = (t-\lambda_{1})^{n_{1}} \dotsb (t-\lambda_{r})^{n_{r}}$$. The usual proof of Jordan-Chevalley decomposition for perfect fields (e.g. [2, Section 4.2]) says that if $$p \in A[t]$$ is any polynomial with $$p-\lambda_{i} \in (t-\lambda_{i})^{n_{i}}$$ for all $$i$$, then $$M_{\mathrm{ss}} = p(M)$$ and $$M_{\mathrm{n}} = M-p(M)$$. (Here often it is assumed that also $$p \in (t)$$, but if I understand correctly it is only used to show that Jordan decomposition is behaves well under passing to $$M$$-invariant subspaces, see part (c) of the Proposition in [2, Section 4.2].) Since by assumption the pairwise differences $$\lambda_{i}-\lambda_{j}$$ are invertible, the polynomials $$(t-\lambda_{i})^{n_{i}}$$ are comaximal in $$A[t]$$, so the Chinese remainder theorem applies and we can find such $$p$$.

If the pairwise differences $$\lambda_{i}-\lambda_{j}$$ are not all invertible, it is not necessarily true that $$M_{\mathrm{ss}},M_{\mathrm{n}} \in \mathrm{Mat}_{n \times n}(A)$$. Here is an example following . Let $$a,b \in A$$ and set $$M_{(a,b)} := \begin{bmatrix} & & a^{2}b \\ 1 & & -(a^{2}+2ab) \\ & 1 & 2a+b \end{bmatrix}$$ which is the companion matrix associated to the polynomial $$(x-a)^{2}(x-b)$$. Then Example 2.3, Example 4.5 of  says that the matrices $$M_{(a,b),\mathrm{ss}},M_{(a,b),\mathrm{n}}$$ have coefficients in $$A$$ if and only if $$a-b$$ is a unit of $$A$$.

Let's work out the example with $$A = \mathbb{Z}$$ and $$a = 0$$ and $$b = 2$$. Then the eigenvalues of $$M_{(0,2)} = \begin{bmatrix} & & \\ 1 & & \\ & 1 & 2 \end{bmatrix}$$ are $$0,2$$, the generalized eigenspace for $$\lambda = 0$$ is $$\{v_{1} = [0,2,-1]^{T} , v_{2} = [2,-1,0]^{T}\}$$ (with $$M_{(0,2)}v_{2} = v_{1}$$) and the eigenspace for $$\lambda = 2$$ is $$\{v_{3} = [0,0,1]^{T}\}$$. Set $$U := [v_{1},v_{2},v_{3}]$$; then $$U^{-1}M_{(0,2)}U = \begin{bmatrix} 0 & 1 & \\ & 0 & \\ & & 2 \end{bmatrix}$$ is a Jordan normal form of $$M_{(0,2)}$$, and $$M_{(0,2),\mathrm{ss}} = U \begin{bmatrix} 0 & & \\ & 0 & \\ & & 2 \end{bmatrix} U^{-1} = \begin{bmatrix} & & \\ & & \\ 1/2 & 1 & 2 \end{bmatrix} \qquad M_{(0,2),\mathrm{n}} = U \begin{bmatrix} & 1 & \\ & & \\ & & \end{bmatrix} U^{-1} = \begin{bmatrix} & & \\ 1 & & \\ -1/2 & & \end{bmatrix}$$ which are not contained in $$\mathrm{Mat}_{3 \times 3}(\mathbb{Z})$$.

References:

 Passi, Roggenkamp, Soriano, "Integral Jordan decomposition of matrices", Linear Algebra and its Applications, vol. 355 (2002), 241 -- 261 (link)

 Humphreys, Introduction to Lie Algebras and Representation Theory, Springer Graduate Texts in Mathematics, vol. 9 (1972)

• Thank you for this quite interesting example. But the whole point of the question is to test the hypothesis mentioned in the first edit of the question. In it, I say that the fact that the characteristic polynomial divides a power of a polynomial with an invertible discriminant in the ring is sufficient. In your example, it is not the case so I do not know yet if my hypothesis is actually sufficient ... or not. – brunoh Oct 8 '18 at 16:52
• By the way I gave you +1 for your effort. This question is old and nobody answered it but if I remember well of my thinking at that time I have absolutely no reference for this hypothesis (hence the question). I just looked carefully at the different proofs of the result in case of a perfect field and I noticed that only this property was really used hence my generalization to the case of a ring. I also remembered that we have to be careful about the generalization of the definitions of separated and semi-simple elements. – brunoh Oct 8 '18 at 20:35
• Okay thanks. Somehow I did not read your edit when I first thought about the question. Please let me know if you find an answer. – Minseon Shin Oct 8 '18 at 20:52