**Question.** Given a positive-definite $n \times n$ matrix $A = (a_{ij})$ with eigenvalues
$$
\lambda_1 \leq \cdots \leq \lambda_n ,
$$
is there a sharp upper bound for the product $\lambda_2 \cdots \lambda_n$ in terms of the quantity
$$
\|A\|_\infty := \max_{1 \leq i, j \leq n} |a_{ij}| ?
$$

A classic inequality due to A. Hirsch states that the modulus of an eigenvalue of an $n \times n$ complex matrix $A$ is less than $n \|A\|_\infty$, which implies that $$ |\lambda_2 \cdots \lambda_n| \leq n^{n-1}\|A\|_\infty^{n-1} . $$ However, this seems like a rather rough estimate for positive-definite matrices. I'm interested in any estimate that is substantially better than this.

**Motivation.** Using Hirsch's inequality one can improve Lemma 4 in page 43 of Siegel's Lectures on Quadratic Forms to yield the following result:

**Theorem.**
If $A = (a_{ij})$ is a positive-definite $n \times n$ matrix, then for every $x \in \mathbb{R}^n$ we have that
$$
\frac{\det(A)}{n^{n-1}a_{11} a_{22} \cdots a_{nn}} \sum a_{ii} x_i^2 \leq \sum a_{ij} x_i x_j \leq n \sum a_{ii} x_i^2 .
$$

The inequality on the left would be greatly improved if we had the sharp upper bound required in the question. This in turn would yield a better answer to this enclosure problem (see my answer to that question).

**Addendum.** If we apply the estimate in Suvrit's answer in the proof of the theorem above, the inequality is indeed improved to:

$$ \frac{\det(A)}{2^{n-1}a_{11} a_{22} \cdots a_{nn}} \sum a_{ii} x_i^2 \leq \sum a_{ij} x_i x_j \leq n \sum a_{ii} x_i^2 . $$

In fact, in the proof the estimate for $\lambda_2 \cdots \lambda_n$ is applied to an auxiliary matrix $B = (b_{ij})$ whose diagonal entries are all $1$ and for which $|b_{ij}| < 1$ if $i \neq j$.

In turn this yields the following improved bound for the enclosure problem:

**Theorem.** Let $E \subset \mathbb{R^n}$ be an $n$-dimensional ellipsoid centered at the origin and containing no other integer point. There exists a transformation $T \in GL(n,\mathbb{Z})$ such that $T(E)$ is contained in the ball of radius
$$
\left(\frac{3}{2}\right)^{(n-1)(n-2)/2} \frac{2^n}{\epsilon_n}\sqrt{2^{n-1}}
$$
centered at the origin.

Here $\epsilon_n$ is the volume of the unit ball of dimension $n$.

*Does anyone know a better bound?*