Just some comments that are well-known in the theory of toric varieties (and no doubt to other areas as well). What we are asked to determine is membership in a finitely generated submonoid $\Gamma$ of $\mathbb N^k$ with more than $k$ generators. The last condition is is a red herring, we can always replace $\mathbb R^k$ by the vector space spanned by the vectors (and the price of possibly replacing $\mathbb N^k$ by some uglier monoid but as we shall see $\mathbb N^k$ will quickly exit the picture).

The first question (which is part of the assumptions of the question) is whether it lies in the subgroup $N$ generated by the same elements. Provided this has a positive answer the next question is whether it lies in the saturation of $\Gamma$, i.e., the submonoid $\Gamma'$ of elements $x$ of $N$ for which $mx\in\Gamma$ for some integer $m>0$. The point about asking this question is that is much easier to answer: The saturation is the intersection of $N$ with the real (or rational) cone spanned by the original vectors. Duality for cones implies that such a cone is the intersection of a finite number of rational half hyperplanes (which can be reasonably efficiently be determined from the original generators, see for instance Ziegler: Lectures on polytopes) and thus that condition can be checked rather easily. Lastly $\Gamma'\setminus\Gamma$ is finite so with a finite number of exceptions membership in $\Gamma'$ implies membership in $\Gamma$. In practice the performing the last step without accepting a finite number of exceptions can be quite tricky as can be seen already in the case when $k$, the rank of $N$, is equal to $1$ (in that case determination of membership in $\Gamma'$ is trivial).

Just some comments that are well-known in the theory of toric varieties (and no doubt to other areas as well). What we are asked to determine is membership in a finitely generated submonoid $\Gamma$ of $\mathbb N^k$ with more than $k$ generators. The last condition is is a red herring, we can always replace $\mathbb R^k$ by the vector space spanned by the vectors (and the price of possibly replacing $\mathbb N^k$ by some uglier monoid but as we shall see $\mathbb N^k$ will quickly exit the picture).

The first question (which is part of the assumptions of the question) is whether it lies in the subgroup $N$ generated by the same elements. Provided this has a positive answer the next question is whether it lies in the saturation of $\Gamma$, i.e., the submonoid $\Gamma'$ of elements $x$ of $N$ for which $mx\in\Gamma$ for some integer $m>0$. The point about asking this question is that is much easier to answer: The saturation is the intersection of $N$ with the real (or rational) cone spanned by the original vectors. Duality for cones implies that such a cone is the intersection of a finite number of rational half hyperplanes (which can be reasonably efficiently be determined from the original generators, see for instance Ziegler: Lectures on polytopes) and thus that condition can be checked rather easily.

[[ Correction: I claimed that $\Gamma'\setminus\Gamma$ is finite which is wrong. ]]

The step from $G'$ to $G$ can also be quite tricky. It is a fact that $G'$ is finitely generated as $G$-module, i.e., there are $x_1,x_2,¼,x_mÎG'$ such that $G'=È_i(G+x_i)$ but the complement $\Gamma'\setminus\Gamma$ may be infinite. An example is given by the monoid generated by $(0,2)$, $(0,3)$ and $(1,0)$ where $(m,1)$ is not in the monoid genated by then but $2(m,1)$ always is.

The relation between $G$ and its saturation can be described in terms of commutative algebra as follows. Consider the monoid algebra $k[G]$, where $k$ is some field. The inclusion $GÍG'$ gives an algebra inclusion $k[G]Ík[G']$ and it makes $k[G']$ the normalisation of $k[G]$ (this gives one way of showing that $G'$ is finitely generated as $G$-module as the same is true for the normalisation). The question of whether $\Gamma'\setminus\Gamma$ is finite or not then has the following interpretation. We have a grading $G®\mathbb N$ of $G$ given by $(m_i)e \sum_im_i$ which induces a grading of $k[G]$ allowing to pass to a projective variety $\mathrm{Proj}k[G]$. Then $\Gamma'\setminus\Gamma$ is finite precisely when $\mathrm{Proj}k[G]$ is normal. The example above was constructed using this; $\mathrm{Proj}k[G]$ is $1$-dimensional with a cusp and hence is not normal.

Just some comments that are well-known in the theory of toric varieties (and no doubt to other areas as well). What we are asked to determine is membership in a finitely generated submonoid $\Gamma$ of $\mathbb N^n$ with fewer than $n$ generators. The last condition is is a red herring, we can always replace $\mathbb R^n$ by the vector space spanned by the vectors (and the price of possibly replacing $\mathbb N^n$ by some uglier monoid but as we shall see $\mathbb N^n$ will quickly exit the picture).

The first question (which is part of the assumptions of the question) is whether it lies in the subgroup $N$ generated by the same elements. Provided this has a positive answer the next question is whether it lies in the saturation of $\Gamma$, i.e., the submonoid $\Gamma'$ of elements $n$ of $N$ for which $mn\in\Gamma$ for some integer $m>0$. The point about asking this question is that is much easier to answer: The saturation is the intersection of $N$ with the real (or rational) cone spanned by the original vectors. Duality for cones implies that such a cone is the intersection of a finite number of rational half hyperplanes (which can be reasonably efficiently be determined from the original generators, see for instance Ziegler: Lectures on polytopes) and thus that condition can be checked rather easily. Lastly $\Gamma'\setminus\Gamma$ is finite so with a finite number of exceptions membership in $\Gamma'$ implies membership in $\Gamma$. In practice the performing the last step without accepting a finite number of exceptions can be quite tricky as can be seen already in the case when $k$, the rank of $N$, is equal to $1$ (in that case determination of membership in $\Gamma'$ is trivial).

Just some comments that are well-known in the theory of toric varieties (and no doubt to other areas as well). What we are asked to determine is membership in a finitely generated submonoid $\Gamma$ of $\mathbb N^k$ with more than $k$ generators. The last condition is is a red herring, we can always replace $\mathbb R^k$ by the vector space spanned by the vectors (and the price of possibly replacing $\mathbb N^k$ by some uglier monoid but as we shall see $\mathbb N^k$ will quickly exit the picture).

The first question (which is part of the assumptions of the question) is whether it lies in the subgroup $N$ generated by the same elements. Provided this has a positive answer the next question is whether it lies in the saturation of $\Gamma$, i.e., the submonoid $\Gamma'$ of elements $x$ of $N$ for which $mx\in\Gamma$ for some integer $m>0$. The point about asking this question is that is much easier to answer: The saturation is the intersection of $N$ with the real (or rational) cone spanned by the original vectors. Duality for cones implies that such a cone is the intersection of a finite number of rational half hyperplanes (which can be reasonably efficiently be determined from the original generators, see for instance Ziegler: Lectures on polytopes) and thus that condition can be checked rather easily. Lastly $\Gamma'\setminus\Gamma$ is finite so with a finite number of exceptions membership in $\Gamma'$ implies membership in $\Gamma$. In practice the performing the last step without accepting a finite number of exceptions can be quite tricky as can be seen already in the case when $k$, the rank of $N$, is equal to $1$ (in that case determination of membership in $\Gamma'$ is trivial).