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I have the following question:

Let A by a TU matrix, then the following matrix is TU $$ \begin{pmatrix}A&&-I\\-1&\cdots&-1\end{pmatrix} $$ i.e. the block matrix composed by $A$, $-I$ (minus the identity matrix) and one final row of $-1$.

I know that in general that statement is false, because the matrix $$\begin{pmatrix}1&-1\\0&1\\-1&-1\end{pmatrix}$$ is not TU and is a submatrix of the matrix $$\begin{pmatrix}1&-1&-1&0\\0&1&0&-1\\-1&-1&-1&-1\end{pmatrix}$$ where $A = \begin{pmatrix}1&-1\\0&1\end{pmatrix}$ is TU.

Nevertheless, I'm wondering if that is true in the case $A$ is the incidence matrix of a bipartite graph (undirected).

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    TU means totally unimodular, right? – 2017-02-12
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    Yes, so for every square submatrix $A'$ of $A$, $det(A') \in \{-1,0,1\}$ – 2017-02-12
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    $\rm A$ is the incidence matrix of a bipartite graph? – 2017-02-12

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