If α i j A i B j = 0 and A i and B j are arbitrary vectors, then prove that α i j = 0This 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?

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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        =  0This 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?

Johnathon Mcmillan · Accepted Answer

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.

rialsv · Accepted Answer

Fix, h,k, then if you take                              A          i                =                  {                                                    1                ,                                                          i                =                h                ,                                                                    0                ,                                                          i                ≠                h                ,                                                                                                  B          j                =                  {                                                    1                ,                                                          j                =                k                ,                                                                    0                ,                                                          j                ≠                k                                                                  then      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.

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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