Inequality for 0-1 matrices

Question

Given an $n \times n$-matrix $A$ with entries only $0$ or $1$ and determinant equal to $\pm 1$. Define the magnitude $M_A$ of $A$ as the sum of all entries of the inverse of $A$.

Question 0: Do we have that the sum of all entries of $A$ is greater than or equal to $M_A$ when $A$ additionally has all diagonal entries equal to one?

This is true for $n \leq 4$. One might also consider other bounds such as whether $M_A \leq n^2$.

Question 1: What is the maximal possible magnitude of such an $A$?

Question 2: What is the maximal value of the magnitude of $A$ in case it has additional all diagonal entries equal to $1$?

This is a small variant of a question asked on Math SE. The zeroth and second question is related to bounds for the magnitude of a Schurian algebra.

Answer 1

The "magnitude" can grow exponentially with $n$, even when $A$ is triangular (and thus has all-$1$ diagonal) with no more than three $1$'s in each row and column.

Indeed suppose $A_{ij} = 1$ if and only if $j \in \{i,i+1,i+3\}$. Then for large odd $n$ the magnitude $M_A$ is asymptotically proportional to $C^n$ for some constant $C>1$, namely the real root $1.46557\ldots$ of $C^3 = C^2 + 1$. For example, $n=99$ yields a matrix with $293$ ones and magnitude $4010964491506511$. (For large even $n$, the magnitude becomes exponentially negative; e.g. $n=100$ yields a matrix with $296$ ones and magnitude $-5878354170831089$.) Exponential growth is the fastest possible for sparse matrices, because each entry of $A^{-1}$ is $\pm \det A'$ for some minor $A'$ of $A$, and Hadamard's inequality gives an exponential upper bound on $\left|\det A'\right|$ for sparse 0-1 matrices $A'$.

The above construction is an adaptation of the example I gave in this Mathoverflow answer of sparse 0-1 matrices that are invertible but have exponentially small eigenvalues, and is the same as the example I gave here of a 0-1 matrix whose inverse has exponentially large absolute sum. The magnitudes for $n=99$ and $n=100$ were computed in milliseconds using this gp code:

S(n, T) = sum(i=1,n,sum(j=1,n,T[i,j])) \\ sum of entries of n*n matrix T
M(n, A,A1) = A = matrix(n,n,i,j,(j==i+1)+(j==i+3))+1; A1 = 1/A; S(n,A1)
M(99)
M(100)

Answer 2

If you allow the entries of $A$ to be in $\{0, \pm 1\}$ then there is a really nice construction that shows that the sum of the entries of $A^{-1}$ can be exponentially large.

Let the diagonals of $A$ be all $1$'s and then make all of the entries underneath the diagonal $-1$ and the entries above the diagonal $0$. Now let $c$ be the first column of $A^{-1}$ and consider what information we get from the equation $A \cdot A^{-1} = I$.

Matrix multiplication tells us that the entries of $c$ satisfy the following recurrence $c_i = \sum_{j < i} c_j$ for $i \ge 2$ and $c_1 = 1$. From this, we easily see that $c_n$ is exponential. Furthermore, the same analysis also holds for the other columns and we get a similar behavior except the $k$th column has $k$ zeros in the beginning before this pattern of the powers of $2$. In conclusion, $A^{-1}$ has all non negative entries and the sum of all the entries of $A^{-1}$ is at least $2^n$.