- 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.
+ The Lyapunov rank of a cone is the dimension of the space of its
+ Lyapunov-like transformations -- that is, the length of a
+ :meth:`lyapunov_like_basis`. Equivalently, the Lyapunov rank is the
+ dimension of the Lie algebra of the automorphism group of the cone.
- 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]_).
+ A nonnegative integer representing the Lyapunov rank of this cone.
+
+ If the ambient space is trivial, the Lyapunov rank will be zero.
+ Otherwise, if the dimension of the ambient vector space is `n`, then
+ the resulting Lyapunov rank will be between `1` and `n` inclusive. A
+ Lyapunov rank of `n-1` is not possible [Orlitzky]_.
- .. [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.
+ .. [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.
+ M. Orlitzky. The Lyapunov rank of an improper cone.
+ http://www.optimization-online.org/DB_HTML/2015/10/5135.html
- .. [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.
+ 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.
sage: positives = Cone([(1,)])
sage: lyapunov_rank(positives)
sage: positives = Cone([(1,)])
sage: lyapunov_rank(positives)
sage: R5 = VectorSpace(QQ, 5)
sage: gs = R5.basis() + [ -r for r in R5.basis() ]
sage: R5 = VectorSpace(QQ, 5)
sage: gs = R5.basis() + [ -r for r in R5.basis() ]
sage: L31 = Cone([(1,0,1), (0,-1,1), (-1,0,1), (0,1,1)])
sage: lyapunov_rank(L31)
1
sage: L31 = Cone([(1,0,1), (0,-1,1), (-1,0,1), (0,1,1)])
sage: lyapunov_rank(L31)
1
sage: L3infty = Cone([(0,1,1), (1,0,1), (0,-1,1), (-1,0,1)])
sage: lyapunov_rank(L3infty)
1
A single ray in `n` dimensions should have Lyapunov rank `n^{2} - n
sage: L3infty = Cone([(0,1,1), (1,0,1), (0,-1,1), (-1,0,1)])
sage: lyapunov_rank(L3infty)
1
A single ray in `n` dimensions should have Lyapunov rank `n^{2} - n
sage: K = Cone([(1,0,0,0,0)])
sage: lyapunov_rank(K)
sage: K = Cone([(1,0,0,0,0)])
sage: lyapunov_rank(K)
sage: e1 = (1,0,0,0,0)
sage: neg_e1 = (-1,0,0,0,0)
sage: e1 = (1,0,0,0,0)
sage: neg_e1 = (-1,0,0,0,0)
sage: L31 = Cone([(1,0,1), (0,-1,1), (-1,0,1), (0,1,1)])
sage: octant = Cone([(1,0,0), (0,1,0), (0,0,1)])
sage: L31 = Cone([(1,0,1), (0,-1,1), (-1,0,1), (0,1,1)])
sage: octant = Cone([(1,0,0), (0,1,0), (0,0,1)])
The cone ``K`` in the following example is isomorphic to the nonnegative
octant in `\mathbb{R}^{3}`::
The cone ``K`` in the following example is isomorphic to the nonnegative
octant in `\mathbb{R}^{3}`::
sage: K = Cone([(2,2,4), (-1,9,0), (2,0,6)])
sage: lyapunov_rank(K) == lyapunov_rank(K.dual())
sage: K = Cone([(2,2,4), (-1,9,0), (2,0,6)])
sage: lyapunov_rank(K) == lyapunov_rank(K.dual())
sage: K1 = random_cone(max_ambient_dim = 8)
sage: A = random_matrix(QQ, K1.lattice_dim(), algorithm='unimodular')
sage: K1 = random_cone(max_ambient_dim = 8)
sage: A = random_matrix(QQ, K1.lattice_dim(), algorithm='unimodular')