- Set A=(aij),u=(u1,u2,u3)⊤. If we use the orthogonality condition (Au,u)=0 with ui=δik we get akk=0. If we use (1) with ui=δik+δim we get
akk+akm+amk+amm=0and hence akm=−amk.
- Set v1=−a23,v2=a13,v3=−a12. Then
Au=(v2u3−v3u2,v3u1−v1u3,v1u2−v2u1)⊤=v×u.
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