NO - Identify a TUM (Totally Unimodular Matrix) (Bad Example)

How can we identify a TU Matrix

• Original Source Youtube

In summary, we have that for each row we have that taken two partition $M_1$ and $M_2$ (which contains some of the elements in that row, taken arbitrarily) and we have that:

• $M_1$ and $M_2$ are not empty
• $sum(M_1)-sum(M_2)=0$

~ Ex.:

This is TUM, because: ✓ All elements are either $-1$, $0$, $+1$ ✓ In each column we have at most 2 non-zero elements ✓ $M_{11}$ = {-1} , $M_{11}$ = {-1} → $sum(M_1)-sum(M_2)=0$ ✓ $M_{11}$ = {0, -1} , $M_{11}$ = {0, -1} → $sum(M_1)-sum(M_2)=0$ ✓ $M_{11}$ = {0} , $M_{11}$ = {0} → $sum(M_1)-sum(M_2)=0$ ✓ $M_{11}$ = {0} , $M_{11}$ = {0} → $sum(M_1)-sum(M_2)=0$

~ Ex.:

This is not TUM, because: ✓ All elements are either $-1$, $0$, $+1$ ✓ In each column we have at most 2 non-zero elements ✗ $M_{11}$ = {1} , $M_{11}$ = {-1} → $sum(M_1)-sum(M_2)=2$ ✓ $M_{11}$ = {1} , $M_{11}$ = {1} → $sum(M_1)-sum(M_2)=0$

NOTE: This goes in conflict with the Sufficient Condition 2 of the video, so probably the $M_1$, $M_2$ condition not as i said.