Generic square matrix with positive 1 bounded entries
Considering a matrix $A=(a_{i.j})$ where $0 \leq a_{i,j} < 1 \forall i,j$. It is important to consider that all entries are strictly lower than 1 and positive.
Rows sum to a number lower than 1
Let us consider that the sum of all entries of matrix $A$'s rows is lower than 1: $\sum_{j=1}^{n}a_{i,j} < 1$. Sorry, maybe I did not specify it, only wrote in the formula, I talk about rows. Rows sum to a number lower than 1.
Determinant…
Let us consider $\det(A)$ (determinant).
Is it true that $\det(A)<1$???
Or maybe $|\det(A)| < 1$???
Best Answer
Note that $\sum_j a_{ij}^2 < \sum_j a_{ij} < 1$. So the magnitude of each row, viewed as a vector in ${\mathbb R}^n$, is less than one. The absolute value of the determinant of $A$ is the volume of the parallelopiped spanned by the rows, which is at most the product of the magnitudes of the row vectors, and therefore is less than one in this case.
If you want to do it algebraically, you can prove it by induction on the dimension, the $1$ by $1$ case being trivial. Then you can do a cofactor expansion along any $i$th row, getting that $$det(A) = \sum_j (-1)^{i + j} a_{ij} \,det(A_{ij})$$ Note that each matrix $A_{ij}$ also satisfies the conditions of the problems, so each $|det(A_{ij})| < 1$ by induction hypothesis. You then get $$|det(A)| < \sum_j |a_{ij}||det(A_{ij})|$$ $$< \sum_j |a_{ij}|$$ $$< 1$$