Question

如何用施密特法把向量组 a1=(1,1,1), a2=(1,2,3), a3=(1,4,9)正交化?

Answer 1

解:
b1=a1=(1,1,1)
b2=a2-(a2,b1)/(b1,b1)b1 = (1,2,3)-(6/3)(1,1,1)=(-1,0,1)
b3=a3-(a3,b2)/(b2,b2)b2-(a3,b1)/(b1,b1)b1
= (1,4,9)-(8/2)(-1,0,1)-(14/3)(1,1,1)
= (1/3,-2/3,1/3).

满意请采纳^_^

Answer 2

b1=a1=(1,1,1)
b2=a2-(a2,b1)/(b1,b1)b1 = (1,2,3)-(6/3)(1,1,1)=(-1,0,1)
b3=a3-(a3,b2)/(b2,b2)b2-(a3,b1)/(b1,b1)b1
= (1,4,9)-(8/2)(-1,0,1)-(14/3)(1,1,1)
= (1/3,-2/3,1/3).