PS: 个人随性整理,对不对,请自行辩证
记 X = [ x 1 , x 2 , … , x i , … , x n ] ∈ R m × n , Z = [ z 1 , z 2 , … , z i , … , z n ] ∈ R n × n X = \big [ x_1,x_2,\dots, x_i, \dots,x_n \big] \in R^{m \times n}, Z=\big [ z_1,z_2,\dots, z_i, \dots,z_n \big] \in R^{n \times n} X=[x1,x2,…,xi,…,xn]∈Rm×n,Z=[z1,z2,…,zi,…,zn]∈Rn×n,
则
Y = X Z = X [ z 1 , z 2 , … ] = [ X z 1 , X z 2 , … ] Y=XZ = X\big [ z_1,z_2,\dots\big]=\big [ Xz_1,Xz_2,\dots\big] Y=XZ=X[z1,z2,…]=[Xz1,Xz2,…]
如果令Y=X,则 z j i z_{ji} zji可以看成是 x j x_j xj 对 x i x_i xi的权重。但前提是X中每个列向量代表一个实例。
记 X = [ x 1 ; x 2 , ; … ; x i ; … ; x n ] ∈ R n × m , Z = [ z 1 ; z 2 ; … ; z i ; … ; z n ] ∈ R m × m X = \big [ x_1;x_2,;\dots; x_i; \dots;x_n \big] \in R^{n \times m}, Z=\big [ z_1;z_2;\dots; z_i; \dots;z_n \big] \in R^{m \times m} X=[x1;x2,;…;xi;…;xn]∈Rn×m,Z=[z1;z2;…;zi;…;zn]∈Rm×m,
则
Y = Z X = [ z 1 ; z 2 ; … ] X = [ z 1 X ; z 2 X ; … ] Y=ZX = \big [ z_1;z_2;\dots\big]X=\big [ z_1X;z_2X;\dots\big] Y=ZX=[z1;z2;…]X=[z1X;z2X;…]
如果令Y=X,则 z j i z_{ji} zji可以看成是 x i x_i xi 对 x j x_j xj的权重。但前提是X中每个行向量代表一个实例。
参考:
Lyu G, Feng S, Huang W, et al. Partial label learning via low-rank representation and label propagation[J]. Soft Computing, 2020, 24: 5165-5176.