# Is the inverse of an invertible totally unimodular matrix also totally unimodular?

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 $$A^{-1}=\det A\cdot \mathrm{adj}(A)=\pm \mathrm{adj}(A).$$

• I think this is true, but I cannot give a link. Nor do I recall the argument. IIRC if you pick a $k\times k$ minor of $A^{-1}$ it is equal to (up to sign and a factor that is a power of $\det A$ - so only sign here) the $(n-k)\times (n-k)$ minor of $A$ formed by the complementary sets of rows and columns. I have a vague recollection that a proof can be found in Jacobson, BA I-II in a chapter on exterior algebra, but cannot check now, as my copy is in my office. Apr 19 '13 at 19:52
• @JyrkiLahtonen: Thank you for your comment. It is very helpful to me. With the tool of exterior algebra I think I can prove your statement now. Apr 20 '13 at 2:43

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.

Here is a more elementary proof.

Theorem: The inverse of a (non-singular) totally unimodular matrix is totally unimodular.

The proof will use the following lemmas.

Lemma 1: Permuting rows and columns preserves total unimodularity.

Lemma 2: Matrix $A$ is totally unimodular if and only if the matrix $$\begin{bmatrix} A & I \\ \end{bmatrix}$$ is totally unimodular (where $I$ is the identity matrix of the appropriate size).

Lemma 3: The pivot operation preserves total unimodularity.

(Pivot operation: choose a non-zero element $a_{i,j}$, called pivot. Multiply the $i$-th row by a scalar so that the pivot becomes 1. Then add multiples of the $i$-th row to all other rows so that all elements in column $j$ except pivot become 0.)

Proof of the theorem:

Let $A$ be a non-singular totally unimodular matrix. By Lemma 2, $\begin{bmatrix} A & I \end{bmatrix}$ is totally unimodular. By repeatedly applying pivot operations we obtain $\begin{bmatrix} I & A^{-1} \end{bmatrix}$ (basically doing Gaussian elimination on $A$). By Lemma 3, this matrix is totally unimodular, and by lemmas 1 and 2, $A^{-1}$ is totally unimodular. $\blacksquare$

Proof of Lemma 1: Let $A$ be totally unimodular, let $\tilde A$ obtained from $A$ by permuting rows and columns and let $\tilde B$ be a square submatrix of $\tilde A$. Then $\tilde B$ is also obtained by permuting some submatrix $B$ of $A$. Permuting rows and columns changes at most the sign of the determinant. Thus $\det(\tilde B) = \pm\det(B) \in \{-1,0,1\}$ $\blacksquare$

Proof of Lemma 2: Let $A$ be totally unimodular. Any square submatrix $B$ of $\begin{bmatrix} A & I\end{bmatrix}$ can be permuted to the form $$\tilde B = \begin{bmatrix} A_1 & 0 \\ A_2 & I_k \\ \end{bmatrix}.$$ where $A_1$ is a square sub-matrix of $A$, and so $\det(A_1) \in \{-1,0,1\}$.

We have

$$\det(B) = \pm\det(\tilde B) = \det(A_1) \in \{-1,0,1\}.$$

On the other hand, if $\begin{bmatrix} A & I\end{bmatrix}$ is totally unimodular, then since deleting columns preserves total unimodularity, $A$ is also totally unimodular. $\blacksquare$

Proof of Lemma 3: Let $A$ be totally unimodular and let $\tilde A$ be obtained from $A$ by the pivot operation with pivot $a_{i,j}$. Let $\tilde B$ be a square submatrix of $\tilde A$. We distinguish three cases:

1. $\tilde B$ contains the pivot row $i$.
Then $\tilde B$ was obtained from some $B$, a square submatrix of $A$, by (i) multiplying a row by $\pm1$ and (ii) adding multiples of that row to other rows. The operations (i) and (ii) preserve the determinant up to sign, so we have $\det(\tilde B) = \pm\det(B) \in \{-1, 0, 1\}$.
2. $\tilde B$ does not contain the pivot row $i$, but contains the pivot column $j$. Since the column of $\tilde B$ corresponding to the pivot is all zeroes, we have $\det(\tilde B) = 0$.
3. $\tilde B$ contains neither the pivot row nor the pivot column. Let $\tilde B_1$ be the matrix obtained from $\tilde B$ by adding the row and column of $\tilde A$ corresponding to the pivot. Since the column of $\tilde B_1$ corresponding to pivot has exactly one 1 and everything else 0, we have $\det(\tilde B) = \pm\det(\tilde B_1)$. But $\tilde B_1$ was obtained from some submatrix $B_1$ of $A$ by operations (as in 1.) that preserve determinant up to sign. So we have $\det(\tilde B) = \pm\det(\tilde B_1) = \pm\det(B_1) \in \{-1,0,1\}$. $\blacksquare$