Writing consecutive ones matrix in reduced row echelon form.

Question

I’m trying to prove that every matrix with the consecutive ones property is totally unimodular.

A matrix has the consecutive ones property if every row is of the form $(0,\ldots ,0,1, \ldots , 1, 0, \ldots, 0) $.

My argument is as follows:

Let M be a matrix with the consecutive ones property. Consider any square submatrix A of M. Note that A also has the consecutive ones property. Consider writing A in reduced row echelon form (RREF). Note that adding multiples of one row to another doesn’t change the determinant, swapping rows only changes the determinant by a factor of -1, and multiplying a row by a scalar changes the determinant by a factor of that scalar. Here, I want to claim something like in a consecutive ones matrix, we would only ever have to multiply a row by a factor of -1 when writing it in RREF. Also note that the resulting RREF matrix will either be the identity, or have at least one row of zeros. In the first case, depending on the row operations used, this means the original submatrix A will have determinant 1 or -1. In the second case, the original submatrix A will have determinant 0. So every nonsingular square submatrix of M has determinant 1 or -1, I.e. M is totally unimodular.

I realised this argument doesn’t work for all binary matrices, as adding a multiple of one row to another may introduce entries other than 0,1 or -1, so we may need to multiply a row by a scalar other than -1. E.g. take the matrix $$ \begin{bmatrix} 1 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{bmatrix} $$

I’m wondering if this argument works at all. Is it true that this doesn’t happen for matrices with the consecutive ones property? Or that if we ever multiply a row by a scalar greater in absolute value than 1, we also divide a row by this scalar? If so, how can I prove this?

@WillJagy sorry, I will edit the post and add the definition — TraKar, Nov 09 '24 at 22:33
When you write $(0,\ldots ,0,1, \ldots , 1, 0, \ldots, 0)$ do you assume that the lists of first zeros or trailing zeros can be empty ? — Jean Marie, Nov 09 '24 at 23:23
@Rob I saw this, but I’m unsure how I can apply it to my argument. Or am I just going down the wrong path altogether by considering RREF? — TraKar, Nov 10 '24 at 00:55
@TraKar, the path is OK; consider Gauss-Jordan Elimination, and while I do have a great answer over at MathOverflow I'll defer to the many more qualified in this subject area. — Rob, Nov 10 '24 at 01:08

score 1 · Accepted Answer · answered Nov 10 '24 at 00:59

First, rearrange the rows of $A$ such that the leading $1$ in each row starts in the same column, or after, as the leading $1$ in the previous row. If two rows have leading $1$’s in the same column, rearrange them such that the one with fewer ones is above the other. Then, among the rows that have the same leading column, subtracting the top-most row from the others still gives a matrix with consecutive ones property. Repeat this procedure until you end up with row echelon form (it doesn’t have to be reduced). Then, either the diagonal entries are all $1$, or there is a zero row. So the determinant of $A$ is either $\pm1$ or $0$.

Writing consecutive ones matrix in reduced row echelon form.

1 Answers1