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For given $n$, the following $n\times n$ complex matrix $M=M^{\dagger}$ is called positive, if

$x^{\dagger}M x\geq 0$

holds for all complex vector $x=(z,z^2,\cdots,z^n)^T$ with arbitrary complex $z$, where $M^{\dagger}$ stands for the conjugate transpose of $M$.

Please note that this definition of positivity is not the same as positive definiteness since $x$ is nor ranging over all vectors here.

Of course, the set of such positive matrices forms a convex cone, I would like to have a closed form of the cone, for a given $N$, how to check whether $N$ is positive?

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    $\begingroup$ a matrix fulfilling your notion of positivity is called "positive definite" in linear algebra - that is, in each linear algebra textbook. you can look for answers there. $\endgroup$
    – Delio Mugnolo
    Commented Sep 13, 2014 at 19:50
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    $\begingroup$ @Delio Please see the constraint on the vector $x$ in our definition. $\endgroup$
    – gondolf
    Commented Sep 13, 2014 at 21:11
  • $\begingroup$ When you say $M^+$ stands for the complex conjugate of $M$, do you mean the conjugate transpose (often written as $M^\dagger$)? $\endgroup$
    – S. Carnahan ♦
    Commented Sep 13, 2014 at 23:32
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    $\begingroup$ "yes of course" - if you mean the conjugate transpose, say the conjugate transpose $\endgroup$
    – Yemon Choi
    Commented Sep 14, 2014 at 12:00
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    $\begingroup$ Moreover, in its present form I find this question frustratingly unmotivated, with no indication of what effort the OP has put into attacking this problem $\endgroup$
    – Yemon Choi
    Commented Sep 14, 2014 at 12:02

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If your question were posed in the real case there is an easy answer. Note that any polynomial $p(\cdot)$ of even degree $2d$ can be written as $p(z)=x^TMx$, where $M$ is a Hermitian matrix and $x=(1,z,z^2,\ldots,z^d)$. In this language your question is how to characterize the cone of nonnegative polynomials (in terms of the matrix $M$): by a classical Theorem of Hilbert, real univariate polynomials are nonnegative if and only if they are sums of squares. That means that $M$ is psd.

Unfortunately, such nice characterization only holds in very few cases for multivariate polynomials (see http://www.msri.org/attachments/workshops/327/553_Lecture-notes_week1_Blekherman.pdf, Theorem 1.2). Moreover, in general, deciding nonnegativity of real polynomials is an NP-hard problem, and therefore one should not expect a closed form description of the cone of such matrices $M$.

Finally, even though I can't figure out a way to describe the complex case within the real one, I suspect there is no nice characterization of this cone.

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  • $\begingroup$ Thank you. Notice here is onyl two variables case when we consider the complex version, so is it possible to find some way to do this? $\endgroup$
    – gondolf
    Commented Sep 14, 2014 at 22:50
  • $\begingroup$ I thought about this, but one cannot simply reduce the complex case to the real one. This is because $i^2=-1$, which does not hold when we write complex numbers as two dimensional vectors. And even if you can do that, in the reference I put above you can see that for bivariate polynomials there are very few degrees where one has nonnegativity iff sum of squares. $\endgroup$
    – Cristóbal Guzmán
    Commented Sep 14, 2014 at 23:03

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