- """
- # First we want to intersect ``K`` with ``W``. The easiest way to
- # do this is via cone intersection, so we turn the subspace ``W``
- # into a cone.
- W_cone = Cone(W.basis() + [-b for b in W.basis()], lattice=K.lattice())
- K = K.intersection(W_cone)
-
- # We've already intersected K with the span of K2, so every
- # generator of K should belong to W now.
- K_W_rays = [ W.coordinate_vector(r) for r in K.rays() ]
-
- L = ToricLattice(W.dimension())
- return Cone(K_W_rays, lattice=L)
-
-
-def lyapunov_rank(K):
- r"""
- Compute the Lyapunov rank (or bilinearity rank) of this cone.
-
- The Lyapunov rank of a cone can be thought of in (mainly) two ways:
-
- 1. The dimension of the Lie algebra of the automorphism group of the
- cone.
-
- 2. The dimension of the linear space of all Lyapunov-like
- transformations on the cone.
-
- INPUT:
-
- A closed, convex polyhedral cone.
-
- OUTPUT:
-
- An integer representing the Lyapunov rank of the cone. If the
- dimension of the ambient vector space is `n`, then the Lyapunov rank
- will be between `1` and `n` inclusive; however a rank of `n-1` is
- not possible (see [Orlitzky/Gowda]_).
-
- ALGORITHM:
-
- The codimension formula from the second reference is used. We find
- all pairs `(x,s)` in the complementarity set of `K` such that `x`
- and `s` are rays of our cone. It is known that these vectors are
- sufficient to apply the codimension formula. Once we have all such
- pairs, we "brute force" the codimension formula by finding all
- linearly-independent `xs^{T}`.
-
- REFERENCES:
-
- .. [Gowda/Tao] M.S. Gowda and J. Tao. On the bilinearity rank of a proper
- cone and Lyapunov-like transformations, Mathematical Programming, 147
- (2014) 155-170.
-
- .. [Orlitzky/Gowda] M. Orlitzky and M. S. Gowda. The Lyapunov Rank of an
- Improper Cone. Work in-progress.
-
- .. [Rudolf et al.] G. Rudolf, N. Noyan, D. Papp, and F. Alizadeh, Bilinear
- optimality constraints for the cone of positive polynomials,
- Mathematical Programming, Series B, 129 (2011) 5-31.
-
- EXAMPLES:
-
- The nonnegative orthant in `\mathbb{R}^{n}` always has rank `n`
- [Rudolf et al.]_::
-
- sage: positives = Cone([(1,)])
- sage: lyapunov_rank(positives)
- 1
- sage: quadrant = Cone([(1,0), (0,1)])
- sage: lyapunov_rank(quadrant)
- 2
- sage: octant = Cone([(1,0,0), (0,1,0), (0,0,1)])
- sage: lyapunov_rank(octant)
- 3
-
- The full space `\mathbb{R}^{n}` has Lyapunov rank `n^{2}`
- [Orlitzky/Gowda]_::
-
- sage: R5 = VectorSpace(QQ, 5)
- sage: gs = R5.basis() + [ -r for r in R5.basis() ]
- sage: K = Cone(gs)
- sage: lyapunov_rank(K)
- 25
-
- The `L^{3}_{1}` cone is known to have a Lyapunov rank of one
- [Rudolf et al.]_::
-
- sage: L31 = Cone([(1,0,1), (0,-1,1), (-1,0,1), (0,1,1)])
- sage: lyapunov_rank(L31)
- 1