The Kronecker delta is
and I use the definition of tensor as
I try to prove the Kronecker delta is a (1,1) tensor. I find the following equation which is enough to show Kronecker delta is a tensor.
$$
{\delta'}_j^i = \delta_k^k \frac{\partial{x_i}'}{\partial{x_k}}\frac{\partial{x_k}}{\partial{x_j}'}
$$
But I dnot know how to get the equation.
Best Answer
This is just chain rule applied to the composition:
$$ (x_1' , \cdots, x_n') \mapsto (x_1 , \cdots, x_n) \mapsto (x_1' , \cdots, x_n')$$
Note that the composition is the identity. So $$ I = \frac{\partial x'}{\partial x} \frac{\partial x}{\partial x'}$$
in terms of index we have \begin{align*} \frac{\partial{x_i}'}{\partial{x_k}}\frac{\partial{x_k}}{\partial{x_j}'}= {\delta'}^i_j \end{align*}