Question

Does anyone have examples of when an object is positive, then it has (or does not have) a square root? Or more generally, can be written as a sum of squares?

Example. A positive integer does not have a square root, but is the sum of at most 4 squares. (Lagrange Theorem). However, a real positive number has a square root.

Another Example. A real quadratic form that is postive definite (or semi-definite) is, after a change of coordinates, a sum of squares. How about rational or integral quadratic forms?

Last Example. A positive definite (or semidefinite) real or complex matrix has a square root. How about rational or integral matrices?

Do you have other examples?

Answer 1

For many examples of this kind, see Olga Taussky, "Sums of squares", Amer. Math. Monthly 77 (1970) 805-830.

Answer 2

An element of $\mathbb{R}[x]$ is a sum of two squares if it is nonnegative as a function on $\mathbb{R}$. This can be seen by noting that its real roots have even multiplicity, its irreducible quadratic factors are of the form $(x-a)^2+b^2$, a product of sums of two squares is a sum of two squares, and a square times a sum of two squares is a sum of two squares.

See Qiaochu's question on Hilbert's 17th problem for what happens in more than one variable.

Answer 3

There is Fejér-Riesz Theorem: a nonnegative trigonometric polynomial can be expressed as the square of the norm of a complex polynomial.

Fejér-Riesz Theorem generalizes from trigonometric polynomials to integrable functions as Szegö’s Theorem.

Answer 4

I think your question lives most naturally in the category of ordered rings.

Here is one example: a field can be ordered iff it is formally real: i.e., iff -1 is not a sum of squares. However, more is true: if x is any element of a field K of characteristic different from 2 which is not a sum of squares, then there exists an ordering < on K in which x is negative. Thus any field which admits more than one ordering will have positive elements which are not sums of squares. For example, in Q(\sqrt{2}), with the usual convention, \sqrt{2} is positive, but it is not a sum of squares, because in a different ordering (here, an adjustment of the given ordering by a field automorphism!) it is negative.

Another Example: No, a positive definite rational or integral quadratic form need not be equivalent to a sum of squares. For instance the quadratic forms x^2 + y^2 and x^2 + 2y^2 are not equivalent over Q. For one thing, the discriminant of the quadratic form (= the product of the coefficients, for a diagonal quadratic form) is well-determined up to a square in the ground field. So it comes back to the fact that in R, every positive number is a square, but not in Q.

For matrices: look at the 1x1 case!

As was alluded to before, another case of this is Hilbert's 17th problem: let K be an ordered field with real closure R. (For simplicity just take K = R = real numbers!) Let f in K(x_1,..,x_n) be a rational function such that for all (a_1,...,a_n) in R^n at which f is defined, f(a_1,...,a_n) >= 0. Then there are rational functions g_1,l..,g_m in K(x_1,...,x_n) such that f = g_1^2 + ... + g_m^2.

Answer 5

I believe another recent question contains an answer:

Hilbert's 17th problem asked if a nonnegative real polynomial is the sum of squares of rational functions. It was answered affirmative by Artin in around 1920.

Answer 6

The existence of the [r, s, n] sum of square formula:

(x1^2+, ...+xr^2)*(y1^2+, ...+ys^2) = (z1^2+, ...+zn^2)

is related to the existence of an axial map of projective spaces:

P^(r-1)xP^(s-1)-->P^(n-1)

There is a recent work extending this formula to some fields of non-zero characteristic:

http://www.uoregon.edu/~ddugger/ksum.pdf

Answer 7

For $x$ a self adjoint element of a $C^*$ algebra it is equivalent:

1. $x$ has non negative spectrum
2. $x$ has a self adjoint square root $x=y^2$
3. $x$ is a finite sum of squares $x=\sum {a_i}^*a_i$

in this case $x$ is indeed called positive.