Proof that $\sum_{i=1}^{i=3} \sum_{j=1}^{j=3} A_{ij} \delta_{ji} = A_{ij}$

Question

Can someone provide a proof that $\displaystyle \sum_{i=1}^{i=3} \displaystyle \sum_{j=1}^{j=3} A_{ij} \delta_{ji} =A_{ij}$? I know that the Kronecker delta is identified with the identity matrix, but I can only make sense of that with vectors, not with matrices.