Simple lower bound for a determinant

Question

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 $A$.

Answer 1

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.

Answer 2

i) If you consider the determinant as a function of the matrix, what exactly you mean by "in terms of the entries of $A$", the determinant is fully specified by the entries in the matrix, so you have more then just a lower bound, you have the exact value if you know the entries... or what you have in mind? For example if $n = 2$ the formula is $$ \det\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} = a_{11}a_{22} - a_{12}a_{21} $$ so maybe you could write $\det(\ldots) \ge \min\{a_{11}, a_{22}\}^2 - \max\{a_{12}, a_{21}\}^2$, does this help you?

ii) If you consider the determinant more a property of the linear transformation associated with the matrix, then it should be the same regardless in what base you express your matrix, so your lower bound should in some sense maybe express this invariant too, do you want to change it if you change the basis?

iii) As the determinant equals the product of the eigenvalues, this could be used to derive some lower bounds maybe, for example if $\lambda_1, \ldots, \lambda_n$ are the eigenvalues of your matrix, then $$ \min_{i=1,\ldots,n} \lambda_i^n \le \det(A) \le n \max_{i=1,\ldots,n} \lambda_i^n. $$