I know that a necessary and sufficient condition for the positivity of a quartic polynomial of many variables is in general difficult. I have a somewhat special case, maybe here more can be said. Let $x\in {\mathbb R}^{n \times m}$ be a real $n\times m$ matrix, $n<m$, and $i=1,\ldots,n$ and $a=1,\ldots,m$. I'm interested in the quartic,

$$Q(x) = \sum_{i,j,a,b,c,d} \;\; x_{ia}\; x_{ib} \; x_{jc} \; x_{jd} \; W_{abcd}$$

where $W_{abcd}\;$ has a number of symmetry properties obviously.

Clearly, if $W_{abcd}\;$, viewed as a quadratic form on symmetric $m\times m$ matrices -- via $W(M) = M_{ab}\; M_{cd}\; W_{abcd}\;$ where $M_{ab}$ is symmetric -- is positive semi-definite, then $Q(x)$ is also non-negative. This is a sufficient condition then for the non-negativeness of $Q$. Were $n=m$ it'd be also necessary.

This requirement is however overly strict because $\sum_i x_{ia}\;x_{ib}$ is not a general $m \times m$ matrix but of rank-n, where we've said $n<m$.

So what would be a necessary and sufficient condition in this case?

  • $\begingroup$ it might help the reader if you explained details of your claim for the case $n=m$. $\endgroup$ – Dima Pasechnik Jul 3 '18 at 16:44

I think $W:=W_{abcd}$ is not well-defined. Indeed, in the case $n=1$ if you want to talk about $W$ as a quadratic form on $m\times m$ matrices, you will have to write $Q(x)=Tr(YY^\top W)=Y^\top WY$, where $Y$ is the vector of all quadratic monomials in $x_{1a}$, $1\leq a\leq m$.

But then in general $W$ is not unique; its entries satisfy linear conditions, one for each coefficient $Q_{ijkl}$ ($i\leq j\leq k\leq l$) of $Q$, but these conditions are too few.

E.g. for $m=3$ your $W$ is $6\times 6$ symmetric matrix, i.e. 21 independent entries, which is less that the number, 15, of the coefficients of $Q$. In fact in this case the nonnegativity of $Q$ is equivalent to the existence of a positive semidefinite (p.s.d.) $W$, as this is nothing but decomposability of $Q$ into a sum of squares of quadratic forms, and the latter is what Hilbert famously proved in 1888. In general, such $W$ is very far from being unique. Such p.s.d. $W$'s form a convex set, 6-dimensional, in general.

That is, you need to fix the question to make it more meaningful.

  • $\begingroup$ Thanks, I'll need to think about it. In any case I'll accept this as the answer as you pointed a serious problem in the question itself. $\endgroup$ – DanielFetchinson Jul 8 '18 at 22:57

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