Linearly independent over $\mathbb{Z}_{\geq 0}$, also linearly independent over $\mathbb{R}_{\geq 0}$? Let $V$ be a set of vectors in $\mathbb{Z}^n$ that are linearly independent over $\mathbb{Z}$. Then they are also linearly independent over $\mathbb{R}$ embedded in $\mathbb{R}^n$, as shown here. My follow up question is:
If $V$ is linearly independent  over $\mathbb{Z}_{\geq 0}$, is it also linearly independent over $\mathbb{R}_{\geq 0}$?
EDIT: What is meant by linearly independent over $\mathbb{Z}_{\geq 0}$ (or $\mathbb{R}_{\geq 0}$ similarly):
Every finite nontrivial linear combination of vectors in $V$ with coefficients in $\mathbb{Z}_{\geq 0}$ must be nonzero.
 A: HINT:
Assume that the system is dependent over $\mathbb{R}_{\ge 0}$. Consider a linear dependence over $\mathbb{R}_{\ge 0}$. If only one coefficient is $\ne 0$, then the corresponding vector is $0$.
Otherwise, we have an equality
$$-v = \sum_{i=1}^m a_i v_i$$
where $v$, $v_i$ are from $V$, and $a_i> 0$.  Now, we can reduce to the case when the system $\{v_i\}_{i=1}^m$ is linearly independent over $\mathbb R$. This is done as following: if the $v_i$ are not linearly independent over $\mathbb R$, consider a linear dependency $\sum b_i v_i = 0$, with some coefficients $b_i>0$. Now substract a convenient multiple of this equality
to get
$$-v = \sum_{i=1}^m (a_i -c b_i) v_i$$
such that all $a_i - c\cdot  b_i\ge 0$, but for some $i$ we have $a_i - c \cdot b_i=0$. Thus we express positively $-v$ in terms of fewer vectors.
Now, we reduced to the case $\{v_i\}_{i=1}^m$ is linearly independent over $\mathbb R$. But then the $a_i$'s are unique, can be determined with the Cramer rule, so they are rational.
$\bf{Added:}$.  The moral of the story: if a system with rational has (positive) real solutions, it also has (positive) rational solutions. In fact
a simple, conceptual argument works. Assume that the system has a real solution.  Now, if a system is compatible ( it has a solution) then the general solution can be expressed with RATIONAL coefficients. So clearly it also has rational solution ( plug in rational numbers into the free variables).  Now, assume that it has a real solution with all components $>0$ ( the ones $=0$ just discard, get a smaller system).  Consider the general solution of the system. Our real solution can be obtained by assigning some values to the parameters. Now, assign rational values, close enough to the last one. We get a rational solution with all entries $>0$.
The fact is that the rational solutions for a compatible system with rational coefficients are dense in the set of real solutions ( that forms a plane of some dimension defined over $\mathbb{Q}$).  Really, that is the crux.
$\bf{Added 2:}$
A finite system of vectors $V$ is linearly independent over $\mathbb{R}_{\ge 0}$ if and only if there exists $u$ in $\mathbb{R}^n$ such that $\langle u, v \rangle >0$ for all $v$ in $V$. Indeed, linear dependence over $\mathbb{R}_{\ge 0}$ means there exists $a_i \ge 0$, $\sum a_i > 0$, and $v_i$ in $V$ such that $0 = \sum a_i v_i$. Dividing the coefficients $a_i$ by their sum, we get $0$ is a convex combination of points in $V$. Therefore, linear independence over $\mathbb{R}_{\ge 0}$ means $0$ is not in the convex hull of $V$. The conclusion follows easily now.
$\bf{Added 3:}$
If $K\subset L$ is an extension of ordered fields, a system of linear inequalities with coefficients in $K$ and having a solution in $L$ will also have a solution in $K$. This follows from the elimination of quantifiers for linear inequalities (Fourier-Motzkin elimination).
A: Here's a similar solution though no Cramer's Rule and a somewhat different feel to it.
You should be able to prove that linear dependence in $\mathbb Z_{\geq 0}$ iff linear dependence over $\mathbb Q_{\geq 0}$.  The next step is to prove $\mathbb R_{\geq 0}$ iff linear dependence over $\mathbb Q_{\geq 0}$. The fact that linear dependence in $\mathbb Q_{\geq 0}$ implies the result in $\mathbb R_{\geq 0}$  is immediate.  Here is the other direction.
suppose you have a linear combination with vectors in $S\subseteq V$ of those vectors and the coefficients are in $x_i \in \mathbb R_{\gt 0}$, the resulting sum is zero (i.e. a linear relation).
Then collect the integer valued column vectors in $S$ in a matrix $A^{(1)}$
Base Case:
if $\dim\ker\big(A^{(1)}\big) = 1$  then up to rescaling, there is only one vector in the kernel which may be chosen strictly positive and rational since $A^{(1)}$ is a rational matrix. You are done.
Recursive Case:
suppose  $\dim\ker\big(A^{(1)}\big) \geq 2$
then working over $\mathbb R$, select $\mathbf x$ and non-zero $\mathbf y\perp \mathbf x$ from $\ker\big(C\big)$.
by orthgonality to a positive vector, $\mathbf y$ has at $j\geq 1$ positive components and $r\geq 1$ negative components.
Then $\mathbf z=\mathbf x + \alpha \mathbf y$ has $r$ negative components for large enough $\alpha \in \mathbb R_{\geq 0}$. These $r$ components are monotone decreasing in $\alpha$ while the others are monotone non-decreasing.  Applying intermediate value theorem (or working through linear equations directly) when
$\mathbf e_i^T\mathbf z = z_i \lt 0 \implies \text{there exists some } \alpha_i \text{ such that  }\mathbf e_i^T\big(\mathbf x + \alpha_i  \mathbf y\big)= 0$
There are finitely many of these $\alpha_i$ so select the smallest (denoting it with the clumsy notation $\alpha^*$) and we find $\mathbf z^* =  \mathbf x + \alpha^* \mathbf y$ is a real non-negative vector with $k\geq 1$ positive components and $m-k\gt 0$ zeros.
This means there is an even smaller subset $S^*$ which has a non-trivial positive real linear relation amongs $k\lt m$ elements.  Collect them in the $k\times k$ matrix $A^{(2)}$ and recurse.
At each step you end up with a strictly smaller matrix that has a positive vector in its kernel when viewed over reals.  The smallest possibility for $A^{(j)}$ is a $1\times 1$ matrix (you can't overshoot to $0 \times 0$, why?). But this algorithm stops iff the Base Case is called one, i.e. iff $\dim\ker\big(A^{(j)}\big) =1$  for some natural number $j$.  At this point the reasoning in the base case applies and we are done.
