On The Symmetry Rank Of Some Symmetric Tensors

The research on the symmetry rank of the symmetry tensor is quite important.The symmetry rank of symmetric tensor can deeply describe the properties of the symmetric tensor.Researching the symmetry rank of the symmetry tensor is helpful to know about tensor.In this thesis,we mainly focus on the research of the symmetry rank problem of the second-order symmetric tensor on arbitrary field and three different special types of the high-order symmetric tensor's symmetry rank problem and the uniqueness of the decomposition.The detail is shown below:Let F be a field,natural number n?2,d?2,e1,e2,…,em?Fn,0?kj?d,k1+k2+…+km=d.we remark:e1k1e2k2…emkm=?ei1(?)ei2(?)…(?)eid.Which,? is the sum of ei1(?)ei2(?)…(?)eid which satisfy all ei1,ei2,…,eid that have kj numbers of ej.From the research for ud-1v,utvd-t,ud-1v+vd three types of symmetric tensors of this thesis,we will obtain:(1)The rank of second order symmetry tensors is all equal to its symmetry rank expect the field with the characteristic of 2's symmetry matrices.On the field with the characteristic of 2,the rank of second order symmetry tensors is equal to its symmetry rank which have non-zero diagonal element,the rank of second order symmetric tensors is equal to its symmetry rank plus one,which diagonal elements are all zero.(2)When char(F)?2 and |F|?d,rs(ud-1v)=d,rs(ud-1v+ud)=d,futher,when xt+1=1 have t+1 different roots on F,we have rs(utvd-t)=t+1.While char(F)=2,we have(?)(3)For the ud-1v,utvd-t,ud-1v+vd three special types of symmetric tensors above,its tensor rank and symmetry rank are identical;symmetry decompositions are not unique.