Determinantal inequality for difference of substochastic matrices

Question

Let $A=(A_{ij})_{1\le i,j\le n}$ be a square matrix with nonnegative real entries. Recall that $A$ is called a substochastic matrix if $$ \forall i,\ \ \sum_j A_{ij}\le 1\ . $$ In the course of my recent work on multidimensional resultants, I was led to the following determinantal inequality $$ \lvert\det(A-B)\rvert\le 1\ , $$ which must hold for every substochastic matrices $A$ and $B$.

My first question is: did this inequality appear previously in the literature?

Note that as Darij said in a comment, the inequality can also be stated as, for every $n\times n$ matrix $M=(M_{ij})_{1\le i,j\le n}$ with real entries $$ \lvert\det(M)\rvert\le \prod_{i=1}^{n}\lVert M_{i\ast}\rVert $$ where $M_{i\ast}$ is the $i$-th row of $M$, and the norm $\lVert x\rVert$ of a vector $x=(x_1,\dotsc,x_n)$ is defined by $$ \lVert x\rVert=\max_{J\subset[n]}\left|\sum_{j\in J}x_j\right|\ . $$ This formulation is reminiscent of the notion of total variation distance between discrete probability measures.

My second question is: can one classify the equality cases, or even better, can one classify extremal pairs of transversals defined in section 4 of the preprint mentioned above?

Answer 1

I guess this type of inequality can be fast proven as follows: Notice that the expression $\det(A-B)$ is an affine function in the entries of $A$ and $B$ considered as variables.

Each row of $A=[a_{i,j}]$ and $B=[b_{i,j}]$ is a vector in $\mathbb{R}^+$ having its entries sum less or equal to $1$. Assume by induction that the claim is true for dimension $n-1$ and we want to prove it for dimension $n$. The inequality is true for $n=1$ or $2$.

Let $n\ge3$, for each $i$, $1\le i\le n$, let $r_i(A)$ be the $i^{th}$ row of $A$ and $r_i(B)$ the $i^{th}$ row of $B$; upon multiplying $A-B$ by a diagonal matrix we may assume that $\max(\sum_{j=1}^na_{i,j},\sum_{j=1}^nb_{i,j})=1$. Also upon exchanging the rows of $A$ and $B$ and multiplying by a diagonal matrix of $\pm 1$ we may assume without loss of generality that $A$ is a stochastic matrix and $B$ a sub stochastic matrix.

Back to $\det(A-B)$, fixing the entries of $A$, to seek the extremum of $\det(A-B)$ we seek the extremum values of each entry in each row of $B$. This is done for either $b_{i,j}=0$ or $\sum_{j=1}^n b_{i,j}=1$. If for any $i$, $\sum_{j=1}^n b_{i,j}=1$ then $\det(A-B)=0$ as $0$ is an eigenvalue, if not there is one row of $B$ say $r_{i_0}$ that is the zero vector. Reconsider $\det(A-B)$ as a function in terms of this row entries from $A$ ($n-1$ variables). Again from the extremums of each entry $a_{i_0,j}=0$ or maximal. Continuing this process we get actually one entry equal to $1$ in this row and the claim follows from the induction hypothesis.