# Positive matrices matrices over commutative rings

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?

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Is the ordering total? Is the ring integral? If so, we can extend the ordering to the quotient field, and then argue as usual (a nonnegative definite matrix can always be written in the form $A^TDT$ where $D$ is a nonnegative diagonal matrix; this does not need diagonalization or any kind of roots). – darij grinberg May 31 '11 at 7:31
Thank you for the remark! Unfortunately, I'd like to answer "no" to the first two questions. I'd like the ordering to be a rather weak one, similar to the "positive cone" of $\ast$-algebras, i.e. elements that can be written as sums of $a^\ast a$ (or just sums of squares with the trivial star involution). This will induce a "ring pre-order". Furthermore, I don't want to restrict myself to integral domains; I have many examples with zero divisors, that I'd like to consider. – Joakim Arnlind May 31 '11 at 7:44

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 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...

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According to his comments, his situation is more general (too general, maybe). – darij grinberg May 31 '11 at 8:43
@darij: yeah, I was too fast. Sorry. So with zero divisors, I believe that this will cause problems indeed. One may have to take a look to the particular situation. :( – Stefan Waldmann May 31 '11 at 8:49
Thanks for the references and the examples Stefan! I have already look them up. Coming back to example no 3 above: I certainly want to have a setup that includes the case of smooth functions on a manifold. Can one find a nice class of rings / algebras (including algebras of smooth functions on a manifold) for which "theorems about positive matrices" hold? What is the crucial property that one needs? Maybe it is technically easier to stick to a positivity defined through positive functionals? – Joakim Arnlind May 31 '11 at 21:54
Dear Joakim, the kind of examples which I have studied a lot are $^*$-algebras over a ring $C = R(i)$ with an ordered ring $R$ in the sense above. For those, you can actually say a lot. And if I'm not completely mistaken, then in the commutative case one might be able to show the positivity of $\mathrm{tr}(AB)$ for positive $A, B$ in this type of examples: maybe even for both canonical positive cones, the sum of square cone and the one defined by positive functionals, which is slightly larger. I will have to think about it... – Stefan Waldmann Jun 1 '11 at 7:56

I tend to believe that the answer to your question, in the generality you want, is negative. Since I am not sure of the proof (and have not written it up in detail), I am making this answer community wiki. Feel free to edit.

We will consider the case $n=2$, and let $R$ be the polynomial ring $\mathbb Z\left[x,y,z,x',y',z'\right]$. Define the nonnegative cone on $R$ to consist of all polynomials which can be written as sums of products of squares and polynomials of the form $a^2x+2aby+b^2z$ with $a,b\in R$ or of the form $a^2x'+2aby'+b^2z'$ with $a,b\in R$.

Let $A$ be the matrix $\left(\begin{array}{cc}x&y\\ y& z\end{array}\right)$, and let $B$ be the matrix $\left(\begin{array}{cc}x'& y'\\ y'& z'\end{array}\right)$. Then, $A$ and $B$ are nonnegative definite (meaning that your condition is satisfied), but I claim that $\mathrm{Tr}\left(AB\right)$ does not lie in the nonnegative cone. Why?

A polynomial in $\mathbb Z\left[x,y,z,x',y',z'\right]$ is said to be positively led if its leading monomial with respect to the lexicographic ordering ($x > y > z > x' > y' > z'$) is a positive integer. The positively led polynomials form a cone, which contains our nonnegative cone because:

(1) the sum and the product of two positively led polynomials are positively led (this is trivial);

(2) squares of polynomials are positively led (this is very easy);

(3) polynomials of the form $a^2x+2aby+b^2z$ with $a,b\in R$ or of the form $a^2x'+2aby'+b^2z'$ with $a,b\in R$ are positively led. (Proving this requires some work. For $a^2x+2aby+b^2z$, we wlog assume that $a$ and $b$ are monomials (because we can always restrict our concentration to the leading monomials of $a$ and $b$; all the other monomials don't contribute anything to the leading monomial of $a^2x+2aby+b^2z$), then show that the monomial $aby$ cannot be $\geq$ to each of the monomials $a^2x$ and $b^2z$ at the same time. Similarly for $a^2x'+2aby'+b^2z'$.)

Now, assume that $\mathrm{Tr}\left(AB\right)$ is nonnegative. Then, $\mathrm{Tr}\left(AB\right) = xx'+2yy'+zz'$ can be written as a sum of products of squares and polynomials of the form $a^2x+2aby+b^2z$ with $a,b\in R$ or of the form $a^2x'+2aby'+b^2z'$ with $a,b\in R$. Now, each of the addends in this sum must have degree $\leq 2$ (where "degree" means "total degree"). This is because the degree of the sum of some positively led polynomials is always equal to the highest of their degrees (and not just smaller or equal to it!), so if we had some terms of degree $\geq 3$, they could not cancel out, contradicting to $\deg\left(xx'+2yy'+zz'\right)=2$.

Now another minor lemma:

(4) If a polynomial of the form $a^2x+2aby+b^2z$ with $a,b\in R$ has degree $\leq 2$, then $a$ and $b$ must be integers (and the polynomial has degree $1$).

This is easy to see by the method we used to prove (3): We assume WLOG that $a$ and $b$ are just monomials (because otherwise, we just replace the polynomials $a$ and $b$ by their leading monomials; this does not change the degree of $a^2x+2aby+b^2z$). Now as in (3) we show that the monomial $aby$ cannot be $\geq$ to each of the monomials $a^2x$ and $b^2z$ at the same time. Hence, the leading monomial in $a^2x+2aby+b^2z$ must be either $a^2x$ or $b^2z$ or $a^2x+b^2z$. In each of these cases, we conclude that at least one of $a$ and $b$ must be an integer (i. e., a monomial of degree $0$), let's say that it's $a$. Now this yields that the leading monomial in $a^2x+2aby+b^2z$ must be $b^2z$, so that $b$ too is an integer. This proves (4).

Similarly:

(5) If a polynomial of the form $a^2x'+2aby'+b^2z'$ with $a,b\in R$ has degree $\leq 2$, then $a$ and $b$ must be integers (and the polynomial has degree $1$).

So let us conclude:

We know that $xx'+2yy'+zz'$ can be written as a sum of products of squares and polynomials of the form $a^2x+2aby+b^2z$ with $a,b\in R$ or of the form $a^2x'+2aby'+b^2z'$ with $a,b\in R$.

But we know that each of the addends has degree $\leq 2$. Thus each of these addends is either a square or the product of two polynomials of the form $a^2x+2aby+b^2z$ with $a,b\in \mathbb Z$ or of the form $a^2x'+2aby'+b^2z'$ with $a,b\in \mathbb Z$. (In fact, any other combination would make the degree too high; in particular, multiplying with a square increases the degree by $\geq 2$ unless the square is just an integer square, and multiplying by a polynomial of the form $a^2x+2aby+b^2z$ with $a,b\in R$ or of the form $a^2x'+2aby'+b^2z'$ with $a,b\in R$ increases the degree by $1$ if $a,b\in\mathbb Z$ or by $\geq 3$ otherwise (due to (4) and (5)).)

We can WLOG assume that all squares occuring in the sum are squares of homogeneous linear polynomials (because we can simply remove the constant terms; here we use that $xx'+2yy'+zz'$ is homogeneous of degree $2$). So we have

(6) $xx'+2yy'+zz' = \left(\text{sum of squares of some homogeneous linear polynomials}\right)$

$+ \sum_{i\in I} \left(a_i^2x+2a_ib_iy+b_i^2z\right)\left(A_i^2x+2A_iB_iy+B_i^2z\right)$

$+ \sum_{j\in J} \left(a_j^2x+2a_jb_jy+b_j^2z\right)\left(A_j^2x'+2A_jB_jy'+B_j^2z'\right)$

$+ \sum_{k\in K} \left(a_k^2x'+2a_kb_ky'+b_k^2z'\right)\left(A_k^2x'+2A_kB_ky'+B_k^2z'\right)$,

where $I$, $J$, $K$ are three disjoint finite sets, and $a_i$, $b_i$, $A_i$, $B_i$, $a_j$, $b_j$, ... are integers.

Now, forget about the lexiographic ordering I introduced (the one that had $x > y > z > x' > y' > z'$), and introduce a new one, with $y > \text{all other variables}$. With the respect to this new ordering, the left hand side of (6) has leading monomial $yy'$. Thus, the right hand side also must have leading monomial $yy'$. Therefore, none of the homogeneous linear polynomials whose squares appear on the right hand side of (6) can contain the variable $y$ (because the square of any such polynomial would contain $y^2$, and thus the leading monomial of the right hand side (6) would be $y^2$ (here we are using again the fact that the degree of the sum of some positively led polynomials is always equal to the highest of their degrees)). But this means that none of the squares on the right hand side (6) can contain the monomial $yy'$ (because such a monomial could only come from a $y$ inside the square, but we have ruled out this possibility). Therefore, the coefficient of $yy'$ on the right hand side of (6) is $\sum_{j\in J}2a_jb_j\cdot 2A_jB_j$. This is divisible by $4$. The coefficient of $yy'$ on the left hand side of (6) is not divisible by $4$. Contradiction.

At least if I didn't mess anything up. Given the length of the proof, this is rather improbable.

Anyway it still keeps the question open whether we are in more luck if we require $R$ to be a $\mathbb Q$-algebra.

EDIT: I think that even if $R$ is supposed to be a $\mathbb Q$-algebra, then your answer is No. Let me try to prove it:

Replace $\mathbb Z$ by $\mathbb Q$, and "integers" by "rationals" throughout the above. We can still get to (6), but we don't get the contradiction through divisibility by $4$ anymore.

Consider (6) again. I have showed that none of the homogeneous linear polynomials whose squares appear on the right hand side of (6) can contain the variable $y$. But the same argument works for any other variable instead of $y$ (just consider the lexicographic order where this variable is higher than all others). This shows that none of the homogeneous linear polynomials whose squares appear on the right hand side of (6) can contain any variables. In other words, these squares are $0$. This simplifies (6) to

(7) $xx'+2yy'+zz' = \sum_{i\in I} \left(a_i^2x+2a_ib_iy+b_i^2z\right)\left(A_i^2x+2A_iB_iy+B_i^2z\right)$

$+ \sum_{j\in J} \left(a_j^2x+2a_jb_jy+b_j^2z\right)\left(A_j^2x'+2A_jB_jy'+B_j^2z'\right)$

$+ \sum_{k\in K} \left(a_k^2x'+2a_kb_ky'+b_k^2z'\right)\left(A_k^2x'+2A_kB_ky'+B_k^2z'\right)$.

Unless the sum $\sum_{i\in I} \left(a_i^2x+2a_ib_iy+b_i^2z\right)\left(A_i^2x+2A_iB_iy+B_i^2z\right)$ on the right hand side of (7) is identically zero, it contributes at least one of the monomials $x^2,xy,xz,y^2,yz,yx,z^2,zx,zy$ with nonzero coefficient to the right hand side of (7), and no other term on the right hand side of (7) can kill this monomial. But this is impossible, as none of these monomials appears on the left hand side of (7) ! Thus, the sum $\sum_{i\in I} \left(a_i^2x+2a_ib_iy+b_i^2z\right)\left(A_i^2x+2A_iB_iy+B_i^2z\right)$ must be identically zero. Similarly, the same holds for the sum $\sum_{k\in K} \left(a_k^2x'+2a_kb_ky'+b_k^2z'\right)\left(A_k^2x'+2A_kB_ky'+B_k^2z'\right)$. Now (7) simplifies to

(8) $xx'+2yy'+zz' = \sum_{j\in J} \left(a_j^2x+2a_jb_jy+b_j^2z\right)\left(A_j^2x'+2A_jB_jy'+B_j^2z'\right)$.

Now, the coefficient of $xz'$ on the right hand side of (8) is $\sum_{j\in J}a_j^2B_j^2$. But the coefficient of $xz'$ on the left hand side of (8) is zero. Thus, $\sum_{j\in J}a_j^2B_j^2=0$. This means that every $j\in J$ satisfies either $a_j=0$ or $B_j=0$. Similarly, every $j\in J$ satisfies either $b_j=0$ or $A_j=0$. Therefore, every $j\in J$ must belong to one of the following pigeonholes:

Pigeonhole 1: $j$'s satisfying $a_j=0\text{ and }b_j=0$.

Pigeonhole 2: $j$'s satisfying $B_j=0\text{ and }b_j=0$.

Pigeonhole 3: $j$'s satisfying $a_j=0\text{ and }A_j=0$.

Pigeonhole 4: $j$'s satisfying $B_j=0\text{ and }A_j=0$.

Any $j$ lying in Pigeonhole 1 can be removed from $J$ without invalidating (8) (because if $j$ lies in Pigeonhole 1, then the addend corresponding to $j$ on the right hand side of (8) is zero and contributes nothing to the sum). Similarly, any $j$ lying in Pigeonhole 4 can be removed from $J$ without invalidating (8). Thus, what remains of (8) is

$xx'+2yy'+zz' = \sum_{j\in \text{Pigeonhole 2}} \left(a_j^2x+2a_jb_jy+b_j^2z\right)\left(A_j^2x'+2A_jB_jy'+B_j^2z'\right)$

$+ \sum_{j\in \text{Pigeonhole 3}} \left(a_j^2x+2a_jb_jy+b_j^2z\right)\left(A_j^2x'+2A_jB_jy'+B_j^2z'\right)$

$= \sum_{j\in \text{Pigeonhole 2}} \left(a_j^2x\right)\left(A_j^2x'\right) + \sum_{j\in \text{Pigeonhole 3}} \left(b_j^2z\right)\left(B_j^2z'\right)$.

This is absurd.

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Thanks for the example. I need to study it in detail when I'm back in my office. I would also expect that the answer to my question is "no" in general. As I wrote in my comment to Stefan's answer, it would be nice to find a class of rings / algebras (including algebras of smooth functions on a manifold) for which the statement always holds. – Joakim Arnlind May 31 '11 at 21:56