例1. 把 α, =(1,1,0,0),αz = (1,0,1, 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)(α2,β,)β, =α2B.0(βr,β)(αs,β)(α3,β,)β, =α3B133(β,β)(β2,β2)(αy,β)(αy,β,)(αs,β3)B.βμ=αs33(β3,β)(β1,β,)(β2,β2)= (1,-1,-1,1)69.2标准正交基区区§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 = −