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Is it possible to prove this matrix family only contains totally unimodular matrices?

The matrix has dimensions $\frac{3n(n-1)}2$ rows and $n+\frac{n(n-1)}2$ columns.

To every pair $(i,i')$ with $1\leq i<i'\leq n$ we have an unique integer $f(i,j)$ from $\big\{n+1,\dots,n+\frac{n(n-1)}2\big\}$ associated with it.

  1. Each $3r+1$ row has three non-zero entries with $M_{(3r+1),i}=M_{(3r+1),i'}=1$ at some $1\leq i<i'\leq n$ (two of first $n$ columns are $1$) and $M_{(3r+1),f(i,i')}=1$.

  2. Each $3r+2$ and $3r+3$ row has two non-zero entries $M_{(3r+2),i}=M_{(3r+2),f(i,i')}=M_{(3r+3),i'}=M_{(3r+3),f(i,i')}=1$.

$$M=\begin{bmatrix}1&1&0&1&0&0\\1&0&0&1&0&0\\0&1&0&1&0&0\\1&0&1&0&1&0\\1&0&0&0&1&0\\0&0&1&0&1&0\\0&1&1&0&0&1\\0&1&0&0&0&1\\0&0&1&0&0&1\end{bmatrix}$$ holds at $n=3$ and it is totally unimodular here.

Is this type of matrix always totally unimodular?

One can prove by induction if we know that if $A,B$ are totally unimodular then $\begin{bmatrix}A\\B\end{bmatrix}$ is totally unimodular under mild non-intersection conditions.

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I don't see that your example matrix $M$ is TU. Taking the first three columns of rows 1, 4 and 7 gives the submatrix $$\begin{pmatrix}1&1&0\\ 1&0&1\\ 0&1&1\end{pmatrix}$$ which has derterminant $-2$.

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  • $\begingroup$ Sorry I wanted to update. I recognized this as well. $\endgroup$ – T.... May 10 '18 at 5:13

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