A linear $1$-form on $\mathbb{R}^n$ is a $1$-form $\alpha=\sum\_i P\_i(x\_1,x\_2,\ldots,x\_n)dx\_i$ such that each $P\_i$ is in the linear form $P\_i=\sum\_j a\_{ij}x\_j$. A linear foliation of $\mathbb{R}^n \setminus \{0\}$ is a foliation tangent to the kernel of a linear $1$-form $\alpha$ whose corresponding matrix $(\alpha\_{ij})$ is a non singular matrix.


As I learned from [this answer](https://mathoverflow.net/questions/282911/a-possible-sub-riemannian-structure-associated-to-a-non-symmetric-matrix/282984#282984), there are linear $1$-forms which are Frobenius integrable but are not closed $1$-form.



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> For $n>2$, is there a complete classification and dynamical description of all linear foliations? In particular, is there a linear foliation of $\mathbb{R}^n \setminus \{0\}$ which has a (compact) leaf with non trivial holonomy? 
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I think that this situation can not occurs when the corresponding $1$-form $\alpha= \sum\_i \sum\_j(a\_{ij}x\_j)dx\_i $ is a closed $1$-form.(Equivalently the matrix $(a\_{ij})$ is a symmetric matrix)



A complete classification of linear foliations of $\mathbb{R}^n \setminus \{0\}$
I don't know offhand the answer of your first question, but I can answer the particular situation you describe afterwards : the holonomy is always trivial.


First, notice that a compact leaf $L$ is everywhere transverse to the radial vector field $R$ (if a ray $x\mathbb R\_{>0}$ were tangent to $L$ then it would be included in $L$, breaking compactness). This implies that $L$ meets any such ray exactly once : it is diffeomorphic to a $(n-1)$-sphere. 


Now, every rescaling $\lambda L$ for $\lambda\in \mathbb R^\times$ is also a compact leaf, thus $\alpha$ foliates the space with smooth spheres. The foliation is therefore smoothly equivalent to the boring trivial fibration $\mathbb R\_{>0}\times\mathbb S^{n-1}$.


