Product of the entries of a matrix

Question

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.

Answer 1

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 $$n\cdot(\prod_{i,j}a_{i,j})^{1/n^2}\leq\rho(A).$$

Answer 2

Not exactly the most frequent mathematical object in the literature, however, here is an interesting instance where this quantity occurs.

Take the the statistical average of $\prod_{i,j} a_{ij}$ over a special unitary $n\times n$ matrix $A$ chosen uniformly at random (i.e., Haar-distributed). Showing this expectation is nonzero for $n$ even is equivalent to the Alon-Tarsi conjecture. For an attempt at explaining why (I think) this conjecture is important, see my answer at this MO post:

What are the current breakthroughs of Geometric Complexity Theory?

Answer 3

Here is another instance of this quantity arising, which is in a similar vein to that of Abdelmalek Abdesselam's answer:

If $A$ is an $n \times n$ real orthogonal matrix, then $\big|\prod_{i,j} a_{i,j}\big| \leq n^{-n^2/2}$. Conversely, equality holds if and only if $A$ is a multiple of a Hadamard matrix (so it is conjectured that equality is attained for some real orthogonal matrix whenever $n$ is a multiple of $4$, but this is of course a long-standing open problem).