西安毛子科技大学XIDIAN UNIVERSITY,αz =(1,0,1,0),例1. 把 α =(1,1,0,0),αg =(-1,0,0,1)α4 =(1,-1,-1,1)变成单位正交的向量组正交化解：令 β =α=(1,1,0,0)(αz,β,β2 =(βi,β)(α3,β)(α, β,β, =α2(β1,β))(β2,β2)(αy,β,)(αy,B(αy,β,)Bβ = αB3(β3,β3)(β1,β)(β2,β,)= (1,-1,-1,1)§9.2 标准正交基 例1. 把 1 2   = = (1,1,0,0), (1,0,1,0), 3 4   = − = − − ( 1,0,0,1) (1, 1, 1,1) 变成单位正交的向量组. 1 1  = = (1,1,0,0) 2 1 2 2 1 1 1 ( , ) ( , )        = − 3 1 3 2 3 3 1 2 1 1 2 2 ( , ) ( , ) ( , ) ( , )             = − − 解：令 4 1 4 2 4 3 4 4 1 2 3 1 1 2 2 3 3 ( , ) ( , ) ( , ) ( , ) ( , ) ( , )                  = − − − = − − (1, 1, 1,1) 正交化 1 1 ( , ,1,0) 2 2 = − 111 ( , , ,1) 333 = −