Say I have a coefficient matrix $M_{m \times n}, n>m$. The only values in the coefficient matrix are $1$s and $0$s.
I have a vector of constants $V_{n \times 1}$ consisting only of positive integers.
If it is shown that the linear system $MX = V$ is consistent, then does that immediately follow that there are integer solutions for $X$? (i.e does solutions in Real Numbers imply integer solutions?)
I have a gut feeling it does, but I can't find a way to prove it.
Also, if the above is fulfilled, and there are no two rows i,j where:
Every single index where $M[i]$ has $1$ is also $1$ in $M[j]$, and $M[j]$ contains $1$s not in $M[i]$, and $V[j] < V[i]$
(i.e The set of indices where $M[i]$ is $1$ is a proper subset of the set of indices where $M[j]$ is $1$ and $V[j]< V[i]$)
e.g :
$$M[i] = (1,1,0,0,0)$$ $$M[j]= (1,1,1,0,0)$$
$$V[i]= 3$$ $$V[j]= 2$$
Does a positive (including zero) integer solution always exist for such a system?
Thanks a lot!