TITLE: Positive definite matrix over a non-Archimedean field
QUESTION [5 upvotes]: Suppose $K$ is a non-Archimedean ordered field, and $A$ is an $n\times n$ square matrix over $K$ such that (1) all diagonal entries of $A$ are $1$, and (2) all off-diagonal entries of $A$ are infinitesimal (i.e. between $-\frac1k$ and $\frac1k$ for all nonzero natural numbers $k$).  For instance, if $K=\mathbb{R}(x)$ is the field of rational functions, ordered with $x$ infinitesimal, then the off-diagonal entries of $A$ could be polynomials in $x$ with zero constant term.
Does it follow that $A$ is positive definite, i.e. that $V^{T} A V > 0$ (in the ordering of $K$) for all nonzero $V\in K^n$?  This seems intuitive to me because it is an "infinitesimal deformation of the identity matrix", but I don't immediately see how to prove it.

REPLY [2 votes]: Here's how to use infinitesimals to make a simple argument. To fix a scale, note that it suffices to prove $V^T A V > 0$ just for those vectors $V$ whose largest component (by absolute value) is $1$.
Let $R$ be the subring  of finite numbers and $M$ the (maximal) ideal of infinitesimal numbers. Note that all matrices and vectors involved have components in $R$.
Define $E = A - I$. Then,
$$ V^T A V = V^T V + (V^T E V) \equiv V^T V \pmod M $$
Since $V^T V \geq 1$, we conclude $V^T A V$ is positive.