I think is precisely . Just expand .
I have to show that if is a tensor then is a linear function of .
My approach:
If I define a function by (einstein summation is used, and ) is the coefficient respect to a basis, then I have:
and
I'm not quite sure about the identity I have used at the , but maybe it follows from:
Let be a basis such that
and
then
by uniqueness of the coefficients i get:
I think I could proove this in the more general case wich I have used where I use a theorem that states that for any tensor (r,s) there are unique coefficients , but there are so much indices so I don't want to wright it here.
, since it is a (1,1) tensor ()is a function on
and you are right, I forgot to write the tr in that expressionBecause how do you evaluate
But you have already solved the problem, by writing
as this is linear for . Same thing holds for more general tensors.
But A is a (1,1) tensor so is a (2,2)
ok so the way I solved it is ok? I'm just starting to read about tensors so I'm easy to confuse.and you are right, I forgot to write the tr in that expression
But you have already solved the problem, by writing
as this is linear for . Same thing holds for more general tensors.
but I wouldn't say that:
is the same as:
I have changed the index. Even though you just mean the coefficients of
the coefficients of is
and I used
I don't know if it's clear what I ment here???