**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](http://www.math.tifr.res.in/~publ/ln/tifr07.pdf) 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](https://mathoverflow.net/questions/87929/ellipsoids-and-lattices-an-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](https://mathoverflow.net/questions/87929/ellipsoids-and-lattices-an-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?*



A spectral inequality for positive-definite matrices 
$\newcommand{\trace}{\operatorname{trace}}$ 
The result below mentions a reasonably improved inequality.


Let $m = \frac{\trace(A)}{n}$, and $s^2= \frac{\trace(A^2)}{n}-m^2$. Then, [Wolkowicz and Styan](http://orion.math.uwaterloo.ca/~hwolkowi/henry/reports/bndseigs80.pdf) (*Linear Algebra and its Applications*, 29:471-508, 1980), show that 


\begin{equation\*}
 \lambda\_1 \ge \frac{\det(A)}{(m+s/\sqrt{n-1})^{n-1}}
\end{equation\*}


*Remark:* As per the notation in the OP, $\lambda\_1$ is the smallest eigenvalue---usually the literature uses $\lambda\_1$ to be largest.


Thus, we obtain the upper bound
\begin{equation\*}
 \lambda\_2\lambda\_3\cdots\lambda\_n \le \left(m+ \frac{s}{\sqrt{n-1}}\right)^{n-1}.
\end{equation\*}
This bound is tight. Consider for example,
If $A= \text{Diag}(1,2,2,\ldots,2)$, then the lhs is $2^{n-1}$, $m=2-1/n$, and $s^2 
= 1/n-1/n^2$, so that $s/\sqrt{n-1} = 1/n$. Thus, the bound on the rhs is tight. 


With $M := \max\_{i,j}|a\_{ij}|$, we see that $m \le M$ and $s^2 \le nM^2 - m^2$, which leads to an upper bound in terms of $M$ as desired


\begin{equation\*}
 \lambda\_2\lambda\_3\cdots\lambda\_n \le \left(M+ \sqrt{\frac{nM^2-m^2}{n-1}}\right)^{n-1} < \left(M + M\sqrt{\frac{n}{n-1}} \right)^{n-1},
\end{equation\*}
which is better than the bound mentioned in the post (though we lost a bit by deleting $m^2$).


