Given a $n \times n$ matrix $A = (a\_{ij})$, I was wondering if there was any theory or research interest relevant to the term


$$ \prod\_{i,j} a\_{ij}$$


the product of all the entries of the matrix.



Product of the entries of a matrix
By using the arithmetic-geometric mean inequality, if each entry $a\_{i,j}$ in $A$ is positive, we can bound several quantities related to $A$ below by the product $\prod\_{i,j}a\_{i,j}$. The geometric-arithmetic mean inequality states that $(x\_1\dots x\_n)^{1/n}\leq\frac{1}{n}(x\_1+\dots+x\_n)$ whenever $x\_1,\dots,x\_n$ are non-negative. Therefore, $n(x\_1\dots x\_n)^{1/n}\leq x\_1+\dots+x\_n$ whenever $x\_1,\dots,x\_n$ are non-negative.


Suppose $A$ is a matrix with non-negative entries. Then we obtain the following bound for the permanent of $A$:


$$\text{per}(A)=\sum\_{\sigma\in S\_{n}}\prod\_{k=1}^{n}a\_{k,\sigma(k)}\geq n!\cdot(\prod\_{\sigma\in S\_{n}}\prod\_{k=1}^{n}a\_{k,\sigma(k)})^{1/n!}=n!\cdot\big((\prod\_{i,j}a\_{i,j})^{(n-1)!}\big)^{1/n!}$$
$$=n!\cdot(\prod\_{i,j}a\_{i,j})^{1/n}.$$


Here, equality is reached if and only if every entry in $A$ is the same. While this inequality is easy to prove, the Van der Waerden's conjecture is a result that was proven in 1980 that strengthens this inequality whenever $A$ is doubly stochastic.


If $A$ is doubly stochastic, then by again applying the geometric-arithmetric mean inequality, we obtain
$\prod\_{i,j}a\_{i,j}\leq n^{-n^2}.$


Van der Waerden's conjecture states that
$$n!\cdot(\prod\_{i,j}a\_{i,j})^{1/n}\leq\frac{n!}{n^n}\leq \text{per}(A)$$
whenever $A$ is doubly stochastic.


For stochastic matrices, the product of all entries can be interpreted in terms of Markov chains.


Observation: Suppose that $(X\_r)\_r$ is an irreducible aperiodic Markov chain with underlying set $\{1,\dots,n\}$ and with transition matrix $A$. Furthermore, suppose that every entry in $B$ is $1/n$.


For almost all tuples $(y\_r)\_r\in\{1,\dots,r\}^{\omega}$, we have
$$\lim\_{N\rightarrow\infty}P(X\_0=y\_0,\dots,X\_N=y\_N)^{1/N}=\prod\_{i,j}a\_{i,j}^{n^{-2}}\leq 1/n.$$


If each entry in $A$ is positive, then the spectral radius $\rho(A)$ of $A$ is an eigenvalue of $A$.


The $i,j$-th entry in $A^{N}$ is the sum of all products of the form
$a\_{i,i\_{1}}\dots a\_{i\_{N-1},j}$. However, the geometric mean value of
$a\_{i,i\_{1}},\dots,a\_{i\_{N-1},j}$ is about $(\prod\_{i,j}a\_{i,j})^{1/n^2}$, so the geometric mean value of the product $a\_{i,i\_{1}}\dots a\_{i\_{N-1},j}$ is about $(\prod\_{i,j}a\_{i,j})^{N/n^2}$. And since the $i,j$-th entry is the sum of $n^{N-1}$ many factors, we estimate that the $i,j$-th entry in $A^N$ is about $n^{N-1}(\prod\_{i,j}a\_{i,j})^{N/n^2}$ which is about
$[n\cdot(\prod\_{i,j}a\_{i,j})^{1/n^2}]^{N}$. Therefore, we have
$$n\cdot(\prod\_{i,j}a\_{i,j})^{1/n^2}\leq\rho(A).$$


