Positive-definite over $\mathbb Q$ form is positive definite over $\mathbb R$?

Question

I was reading P. Etingof's "Introduction to the representation theory" when I found this problem (and I've trouble with it): we have a quadratic form $Q(x) = \sum x_i^2 - \frac{1}{2}\sum b_{ij}x_ix_j$, $b_{ij} \in \mathbb N$ (if it can be important, $b_{ij}$ is an adjacency matrix of some graph) and $Q$ is positive-definite over $\mathbb Q$.

Now our goal is to prove that $Q$ is positive-definite over $\mathbb R$.

It's obvious that it's at least positive-semidefinite (Etingof's hint is to use that $\mathbb Q$ is dense in $\mathbb R$ and here it works). But why positive-definite?

My idea was to use the Sylwester criterion two times... But is it true for rational numbers?

Answer 1

Quadratic forms can be diagonalized over any field of characteristic not 2. So $Q$ has a diagonalization $\langle a_1,\dots,a_r\rangle$ with $a_i\in \mathbb{Q}$, and since $Q$ is positive definite, $a_i>0$. Now diagonalizations are compatible with base change, so this is still a diagonalization over $\mathbb{R}$, and $Q_\mathbb{R}$ is still positive definite.

Answer 2

Suppose this matrix is positive semidefinite over $\mathbb{R}$, but not positive definite. Then it has a zero eigenvalue. Now show that at least one eigenvector corresponding to this eigenvalue has rational coordinates, since the form itself has rational elements. This implies that this matrix is also not positive definite over $\mathbb{Q}$.

Answer 3

The map $Q: \mathbb{R}^n \to \mathbb{R}$ is continuous, and for every nonzero rational $v$, we have that $Q(v) > 0$. Hence around every rational nonzero $v$, there is a neighbourhood where $Q > 0$. Every nonzero real $v$ falls into one of these neighbourhoods.