If alpha_(ij) A^i B^j=0 and A^i and B^j are arbitrary vectors, then prove that α_(ij)=0



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If α i j A i B j = 0 and A i and B j are arbitrary vectors, then prove that α i j = 0
This problem appeared in my Differential Geometry class, the professor explained the problem by, first taking an arbitrary vector and demonstrating that α i i = 0. and then proceeded to demonstrate that, A l = B m = 1 , ( 1 l n , 1 m n , l m ). I get the proof somewhat. Can any of you elucidate it or give an alternative proof?

Answer & Explanation

Johnathon Mcmillan

Johnathon Mcmillan

Beginner2022-10-09Added 7 answers

A i B j are the component of the tensor product of two vector A B . Among all these tensors there are also the tensors e i e j , where B = { e 1 , , e n } is a base of the vector space V,and B = { e i e j ,   i , j = 1 , , n } is a base of the space of rank 2 tensors over tensors over V.
α i , j A i B j is the inner product of the tensor α and the tensor A B
α i , j A i B j = α : ( A B )
You know that if the inner product of a vector for each element of a base vanish, then the vector is the null vector. This is true also for the inner product vector space of tensors.


Beginner2022-10-10Added 3 answers

Fix, h,k, then if you take
A i = { 1 , i = h , 0 , i h , B j = { 1 , j = k , 0 , j k
α i j A i B j = 0 α h k = 0 ,
for the arbitrariness of h,k, this is true for all h,k.

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