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  1. Diagonally dominant matrices are known to be positive semidefinite (PSD). What are some other families of (symmetric) matrices which are known to be PSD?

  2. Given an arbitrary matrix A, the positivstellensatz or 'sum of squares' machinery can be used as a tool towards showing that A is PSD. What are other simpler methods/tools that can be used towards proving that a matrix is PSD?

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  • $\begingroup$ Note that $A$ is PSD iff $(A+A^T)/2$ is PSD, since $x^T A x = x^T\frac{A+A^T}{2} x$. So in (2) you can assume that $A$ is symmetric. $\endgroup$ – Yuval Filmus Dec 9 '14 at 6:26
  • $\begingroup$ This seems to be two discrete questions and should be asked as such. $\endgroup$ – David Richerby Dec 9 '14 at 7:27
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To get the obvious out of the way first, to show that the $n\times n$ matrix $M$ is PSD, it is enough to show that it is the Gram matrix of $n$ vectors, i.e. $m_{ij} = \langle u_i, u_j\rangle$ where $u_1, \ldots, u_n \in \mathbb{R}^n$. (This is a "sum of squares" proof of positive semi-definiteness, because it shows that the bi-linear form $x^T M x$ can be written as a sum of squares of linear functions of $x$. I am wondering if you meant something more sophisticated by "sum of squares".)

You get a useful class of PSD matrices from positive definite functions. A function $f: \mathbb{R}^d \to \mathbb{R}$ is positive definite if for all sequences $u_1, \ldots, u_n \in \mathbb{R}^d$, the $n\times n$ matrix $M$ defined by $m_{ij} = f(u_i - u_j)$ is PSD. It is not very hard to see that if $f$ is the Fourier transform of a finite measure, then $f$ is positive definite. I.e. it is sufficient that there is a finite Borel measure $\mu$ on $\mathbb{R}^d$ such that for all $x$ $$ f(x) = \int{e^{i\langle x, y\rangle} d\mu(y)}. $$ A classical theorem of Bochner shows that this condition is necessary as well.

A special class of positive definite functions are the functions $f(x) = g(\|x\|_2)$, for which the matrix $M$ becomes $m_{ij} = g(\|u_i - u_j\|_2)$. Another classical result of Schoenberg shows that such an $f$ is positive definite if and only if there exists a finite Borel measure $\mu$ on $[0, \infty)$ so that for every $t \geq 0$ $$ g(t) = \int_{0}^\infty{e^{-t^2s}d\mu(s)}. $$

The original motivation for Schoenberg's theorem was to determine what functions $g$ have the property that for any $u_1, \ldots, u_n \in \mathbb{R}^d$, the metric space with metric $d(u_i, u_j) = g(\|u_i - u_j\|_2)$ embeds isometrically into Euclidean space.

This blog post has a simple proof of Schoenberg's theorem and many links.

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