If $A$ was a totally unimodular matrix, then show the matrix obtained by augmenting with the identity matrix $[A|I]$ is TU. Also show the matrix obtained by augmenting with the zero matrix matrix [$A|0]$ is TU.
I know TU matrix implies $A$ has to have all entries of $[0,1,-1]$. Does augmenting mean elementary row operation? If so, can any one provide some brief hints on how to solve this?
Let $A$ be a totally unimodular matrix.
Consider appending a single column of standard unit basis to a totally unimodular matrix. Let $B= [A, e_i]$.
Now, let's consider a submatrix of $B$. If it doesn't involve the last column, the determinant is clearly $0,1$, or $-1$.
If it involves the last coumn, then let's perform Laplace expansion along the last column, it will be either be $0$ or $c$ times determiant of submatrix of $A$ where $c \in \{1,-1\}$.
Hence appending a single column consisting of standard unit basis preserve the totally unimodular property.