Let $A$ in $\mathbb{Q}^{n \times n}$ such that $\det(A) > 0$? Is there a simple lower bound for $\det(A)$ in terms of the entries of $A$?
Edit: Motivation: Let $M$ be an $m \times n$ matrix. I want to compute a lower bound for
$$
\frac{\det(A^*_i)}{\det(A_i)},
$$
where $\det(A_i)$ is a $n \times n$ matrix consisting of columns of $M$ and $A^*_i$ equals $A_i$ where the $i$-th column is replaced by some column vector $b$ (Cramer's Rule). I know that $\det(A_i)$ can be upper bounded by $2 \langle M \rangle$, where $\langle M \rangle$ denotes the encoding length of $M$. The encoding size of the lower bound should be bounded polynomially in $n$ and the encoding size of $A$.
Best Answer
I would say "sort of". If the entries of $A$ are all integers, then $\det A$ is an integer, so if $\det A>0$ we must have $\det A\ge 1$. This can be generalised to if the entries of $A$ are all rational - multiply $A$ by a positive integer $k$ such that each entry of $kA$ is an integer. It is not hard to see then that $\det A\ge\frac1{k^n}$. If $A$ has arbitrary real or complex entries then of course $A$ can have arbitrarily small determinant.