0
$\begingroup$

Given a matrix $M$, we will refer to the submatrix formed by the first $k$ rows as $M([k], \cdot)$.

Let $A$ be a $m\times n$ totally unimodular matrix where $m \leq n$. We define a new $m\times (n+1)$ matrix $A'$ by appending onto $A$ a column $(v_1, v_2, \dots, v_m)^T$, where $v_1, v_2, \dots, v_{m-1}$ are known and $v_m$ is unknown, such that $A'([m-1], \cdot)$ is totally unimodular. Can we always find a value for $v_m$ such that $A'$ is totally unimodular?

Computations for small size matrices indicate that we can, but attempting to prove this has not been fruitful.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

No, here is a counterexample with $m=n=4$, found using a computer: $$\begin{pmatrix} 1 & 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 & v_4 \end{pmatrix}$$ The submatrices obtained by discarding the last column and last row respectively are totally unimodular by the consecutive-ones property (this implication was noted in Section 8 of Delbert Ray Fulkerson and Oliver Alfred Gross, “Incidence Matrices and Interval Graphs”, Pacific Journal of Mathematics 15 (1965), 835–855), but the determinant of the $4\times4$ submatrix obtained by discarding the fourth column is equal to $v_4-3$. Since $v_4\in\{-1,0,1\}$, this determinant is in $\{-4,-3,-2\}$, violating total unimodularity.

Improve this answer
$\endgroup$
3
  • $\begingroup$ That's a nice counterexample, thank you for that! I wonder if a similar counterexample exists if we require that the starting matrix $A$ has full rank. $\endgroup$ Commented Jul 18 at 9:41
  • 1
    $\begingroup$ @KevinS. Changing $A_{44}$ from $0$ to $1$ in the example I gave should work. $\endgroup$ Commented Jul 18 at 10:05
  • $\begingroup$ Just checked it and you're right, that does work! $\endgroup$ Commented Jul 18 at 10:48

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.