(1)求H1,作A2=H1A。 1=sgn(a1)∑a2)2=3 a,to 2,u=(4,2,2) 3×4=12: 1-2-2 H1=I-p1u=-22-1(1) 求 H1 ，作 A2 = H1 A。 1 sign ( )( ) 3; 2 3 1 1 2 1 = 11  1 = i=  a ai  2 4, 2, (4,2,2) ; 1 11 1 2 T u = a + = u = u =  3 3 4 12; 1 =1 u1 =  =            = − = − 3 1 2 2 2 1 1 2 2 3 1 1 1 1 - - - - - - - H I uu T 