%0 Journal Article
%T Rank 1 Decompositions of Symmetric Tensors Outside a Fixed Support
%A E. Ballico
%J ISRN Geometry
%D 2013
%R 10.1155/2013/704072
%X Let , , denote the degree Veronese embedding of . For any , let be the minimal cardinality of such that . Identifying with a homogeneous polynomial (or a symmetric tensor), corresponds to writing as a sum of powers with a linear form (or as a sum of -powers of vectors). Here we fix an integral variety and and study a similar decomposition with and minimal. For instance, if is a linear subspace, then we prove that and classify all such that . 1. Introduction Let , , denote the degree Veronese embedding of . Set . For any , the symmetric rank or symmetric tensor rank or the rank of is the minimal cardinality of a finite set such that , where denotes the linear span. In this paper we study the following problem. Fix an integral variety . What is the minimal cardinality of a finite set such that , for any and ? The -rank of is the minimal cardinality of a finite set such that , with the convention if there is no such set , that is, if . The classical case is when is a proper linear subspace , , of . Fix any -dimensional linear subspace and . In this case we are looking at the symmetric rank of a symmetric tensor depending only on variables. By [1, Proposition 3.1], there is such that and ; that is, . By [2, Exercise ], we have ; that is, if , , and , then . The general case is motivated from the following query. Assume that writing the -power of the linear form associated with has a price . Find a finite set such that and is small. What happens if the points of are cheaper than the points of ? The condition ¡° for any ¡± is necessary to get a reasonable definition for the following reason. Suppose ; that is, , and take such that and . Fix any . Obviously . Hence there is a set such that , , and . We need to exclude sets like . A similar problem is to find the minimal cardinality of a set such that . It is easy to see that is always a finite integer. Let be the minimal integer among all finite sets such that , for any and . Obviously . Roughly speaking, we use if it is forbidden (or very expensive) to use the points of . We first prove the following result. Proposition 1. Fix any and any .(a)One£¿£¿has£¿£¿ .(b)If£¿£¿ , £¿£¿then£¿£¿ £¿£¿and£¿£¿every£¿£¿ £¿£¿evincing£¿£¿ £¿£¿evinces£¿£¿ . £¿£¿If , £¿£¿then£¿£¿ . Quite often equality holds in part (a) of Proposition 1 (see, for instance, Theorem 6). Part (b) shows that if , then is always small. We recall again that if is a linear subspace, then for every ([2, Exercise ]). In this paper we prove that this is a characterization of linear subspaces of . Indeed we prove the following result. Theorem 2. Fix an integral variety which is not a
%U http://www.hindawi.com/journals/isrn.geometry/2013/704072/