Assume that $R$ is a commutative ring with a ring compatible ordering and let $A$ and $B$ be symmetric $n\times n$ matrices with entries in $R$ such that $\sum x\_iA\_{ij}x\_j\geq 0$ and $\sum x\_iB\_{ij}x\_j\geq 0$ for all $x=(x\_1,\ldots,x\_n)\in R^n$. Is it true that $\operatorname{tr}(AB)=\sum A\_{ij}B\_{ji}\geq 0$?


This is true for matrices, and the proof usually involves the Kronecker product and diagonalization of the matrices, which is problematic over rings (well, not the Kronecker product of course). Does anyone know how to find a theorem like this for rings?
In general, I wonder how much of the standard properties of positive matrices that can be extended to rings?



Positive matrices matrices over commutative rings
Hi Joakim.


Just to clarify the notions: an ordered ring is a commutative ring $R$ (say with $1 \ne 0$) with a distinguished subset $P$, called the positive elements, such that $P + P \subseteq P$ and $P \cdot P \subseteq P$ and $R = -P \cup \{0\} \cup P$ is a *disjoint* union. Then one defines $a \le b$ iff $b-a \in P$. It follows that $1 > 0$ and $R$ has characteristic $0$ and not zero divisors. Is this the type of ordering you're interested? I hope so, in this case, one can apply the following stuff:


Then there are different notions of positivity for matrices: First it is convenient to pass to the complex version, i.e. to a ring extension $C = R(i)$ with $i^2 = -1$. Then the matrices $M\_n(C)$ become a $^\*$-algebra over $C$ with the usual transposition and complex-conjugation of matrices.


The most intrinsic definition is then to say that $A \in M\_n(C)$ is positive iff $\omega(A) \ge 0$ for every positive functional $\omega\colon M\_n(C) \longrightarrow C$. Here a positive functional is a $C$-linear functional with $\omega(A^\ast A) \ge 0$, as you define that in arbiotrary $^\*$-algebras over $\mathbb{C}$.


It is then a theorem that the following equivalences hold:


1. $A$ is positive
2. $\langle z, Az\rangle \ge 0$ for all $z \in C^n$


Necessarily $A^\* = A$ for a positive matrix. Moreover, all positive functionals $\omega$ are of the form $\omega(A) = \mathrm{tr}(\varrho A)$ with a positive matrix $\varrho$.


It is then clear that $\mathrm{tr}(AB) \ge 0$ for two positive matrices...


So the situation is very much parallel to the case $C = \mathbb{C}$. You can find proofs of this in an appendix of a paper of Henrique Bursztyn and mine (Henrique Bursztyn, Stefan Waldmann: Algebraic Rieffel Induction, Formal Morita Equivalence and Applications to Deformation Quantization. J. Geom. Phys. 37 (2001), 307-364). In fact, over the last years, we studied a lot the representation theories of $^\*$-algebras over ordered rings, so this might be interesting for you.


EDIT: according to the comments, this is not really the situation you're interested in. Sorry. But I guess one can still learn something from it. For the more general situation of a convex cone in $R$, defining the positivity, I can not say much but present some examples:


1.) For commutative $C^\ast$-algebras, i.e. continuous functions on compact Hausdorff spaces $X$, the standard positive cone is that defined via the positive functionals and it coincides with simply every other reasonable definition of "postivity". There you do have the property of the trace, as you can determine positivity pointwise in $X$. But I guess, you have this as a well-known motivating example anyway... :)


2.) For $O^\ast$-algebras the situation is already very complicated. There are different notions of positivity defined by various cones. The one via positive functionals is still very canonical, but most of the time not soo useful. I guess a closer look at Schmuedgen's book on $O^\*$-algebras will provide a lot of details and examples.


3.) For, say smooth, functions on a manifold, I have just recently learned that there are two characterizations of positivity: you can ask for the cone generated by squares or the one via positive functionals. Clearly the later contains the first. But in general, this inclusion is proper, see this question [Hilbert's 17th Problem for smooth functions](https://mathoverflow.net/questions/53876/hilberts-17th-problem-for-smooth-functions) and David's answer. So in this case, the characterization via positive functionals gives the pointwise postivity (you need to show that a positive functional is automatically continuous in the sup-norm and hence a positive Borel measure). Hence for this cone, your trace property holds as well (being able to check it pointwise).


OK, these are the examples I have at hand at the moment, maybe I get some more...


