The book I am reading asks the reader to verify that the Kronecker Delta is a second-order mixed tensor with one contravariant and one covariant index as indicated:
$$ \delta_j^i = \left\{\begin{array}{ll} 1 & i = j\\ 0 & i \neq j \end{array} \right.$$
By definition, if it were such a tensor we would have:
$$ {\delta'}_j^i = \delta_l^k \frac{\partial{x_i}'}{\partial{x_k}}\frac{\partial{x_l}}{\partial{x_j}'}$$
using the summation convention. I have seen a "proof" of this equation somewhere else that goes on to say:
$$ {\delta'}_j^i = \delta_k^k \frac{\partial{x_i}'}{\partial{x_k}}\frac{\partial{x_k}}{\partial{x_j}'} = \frac{\partial{x_i}'}{\partial{x_j}'} = \left\{\begin{array}{ll} 1 & i' = j'\\ 0 & i' \neq j' \end{array} \right.$$
supposedly verifying it by using the fact that ${\delta}_l^k = 0$ if $ k \neq l$.
However, since k is a dummy index that gets summed over, shouldn't we have $$ {\delta'}_j^i = \left\{\begin{array}{ll} n & i' = j'\\ 0 & i' \neq j' \end{array} \right.$$
where n is the dimension of the space?
For example, if i' = 1, j' = 1, n = 2, wouldn't the transformation equation above give:
$$ {\delta'}_1^1 = \delta_1^1 \frac{\partial{x_1}'}{\partial{x_1}}\frac{\partial{x_1}}{\partial{x_1}'} + \delta_2^2 \frac{\partial{x_1}'}{\partial{x_2}}\frac{\partial{x_2}}{\partial{x_1}'} = 2 $$
where I've used $\delta_1^2 = \delta_2^1 = 0$? Where am I going wrong here?
