My question is learned from here. Let me restate it as follows:
A unimodular matrix $M$ is a square integer matrix having determinant $+1$ or $−1$. A totally unimodular matrix (TU matrix) is a matrix for which every square non-singular submatrix is unimodular.
Now suppose an $n\times n$ non-singular matrix $A$ is totally unimodular. Can we prove that $A^{−1}$ is also totally unimodular? Or if it is not correct, can we have a counterexample? Any help is much appreciated.
Edit: What I have already known is that the statement is true when $n=2$ and $3$. It follows from the definition of unimodular matrix and the fact that if $A$ is unimodular, then $$A^{-1}=\det A\cdot \mathrm{adj}(A)=\pm \mathrm{adj}(A).$$
Best Answer
As commented by Jyrki Lahtonen, the statement is true and it is immediately implied by the following relation between the minors of $A^{-1}$ and the minors of $A$.
Proposition: If $A$ is an invertible $n\times n$ matrix, and if $i_1,\dots,i_n$ and $j_1,\dots,j_n$ be two permutations of $1,\dots,n$, then the minor of $A$ corresponding to rows $i_1,\dots,i_k$ and columns $j_1,\dots,j_k$, denoted by $d$, and the minor of $A^{-1}$ corresponding to rows $j_{k+1},\dots,j_n$ and columns $i_{k+1},\dots,i_n$, denoted by $d'$,satisfy that $$d=\pm d'\det A.$$
Proof: Let $e_1,\dots,e_n$ be a basis of $\mathbb{R}^n$, and let $f_i= A e_i$, $i=1,\dots,n$. Then on the one hand, $$\omega:=Ae_{j_1}\wedge\cdots\wedge Ae_{j_k}\wedge e_{i_{k+1}}\wedge \cdots\wedge e_{i_n}=\pm d\cdot e_1\wedge\cdots\wedge e_n,$$ on the other hand, $$\omega=f_{j_1}\wedge\cdots\wedge f_{j_k}\wedge A^{-1}f_{i_{k+1}}\wedge \cdots\wedge A^{-1}f_{i_n}=\pm d'\cdot f_1\wedge\cdots\wedge f_n.$$ Since $$f_1\wedge\cdots\wedge f_n=\det A\cdot e_1\wedge\cdots\wedge e_n,$$ the conclusion follows.