For an non-negative-definite symmetric bilinear form on $\Bbb Z^n$ there is a basis $B$ of $\Bbb Z^n$ with $v\cdot v\geq 0$ for all $v\in B$

Question

Let $A:\Bbb Z^n\times \Bbb Z^n\to \Bbb Z$ be a symmetric bilinear form. I want to show that there exists a basis $\{v_1,\dots,v_n\}$ of $\Bbb Z^n$ such that $A(v_i,v_i)\geq 0$ for all $i=1,\dots,n$. Of course this is false when $A$ is negative definite, so assume this is not the case. Thus, there is at least one $v\in \Bbb Z^n$ such that $A(v,v)\geq 0$. We may assume $v$ is not an integer multiple of another $w\in \Bbb Z^n$ with $w\neq \pm v$, so that we can extend $\{v\}$ to a basis of $\Bbb Z^n$. But then I can't see how to proceed. Any hints?

Answer 1

Hints:

• Given that $\ A(v,v)>0\ $, then for any other $\ v^\prime\in\mathbb{Z}^n\ $ $$ A\left(zv+v^\prime,zv+v^\prime\right)=z^2A(v,v)+2zA\left(v,v^\prime\right)+A\left(v^\prime v^\prime\right)>0 $$ for sufficiently large $\ z\in\mathbb{Z}\ $.
• If $\ \left\{v, v_1, v_2,\dots,v_{n-1}\right\}\ $ is a basis of $\ \mathbb{Z}^n\ $, then so is $\ \left\{v, v_1+z_1v, v_2 +z_2v,\dots,v_{n-1} +z_{n-1}v\right\}\ $ for any $\ z_1,z_2,\dots,z_{n-1}\in\mathbb{Z}\ $.