I am trying to simplify a tensor equation with Kronecker delta
$$ A_{ij} \big ( \delta_{ik}\delta_{jm} -\frac{1}{3}\delta_{ij}\delta_{km} \big) $$
$A$ and $\delta$ are Cartesian tensors.
I know definition of Kronecker delta
$$\delta_{ij}=
\begin {cases}
0 \quad \text {if} \quad i \ne j \\
1 \quad \text {if} \quad i = j
\end {cases}
$$
and $\delta_{ii} = 3$
But I am confused how to simplify the equation. The answer at the end of simplification seems $A_{km}$. Any direction or comments?
Best Answer
Let's write sums, just for now: $$\begin{align} \sum_{i,j} A_{ij}\big( \delta_{ik}\delta_{jm} -\frac{1}{3}\delta_{ij}\delta_{km} \big) &= \sum_{i,j}A_{ij}\delta_{ik}\delta_{jm} - \frac{1}{3}\sum_{i,j}A_{ij}\delta_{ij}\delta_{km} \\ &= \sum_{i,j}A_{ij}\delta_{ik}\delta_{jm} - \frac{1}{3}\delta_{km}\sum_{i,j}A_{ij}\delta_{ij} \\ &= A_{km} - \frac{1}{3}\delta_{km}(A_{11}+A_{22}+A_{33}),\end{align}$$so the answer is not only $A_{km}$.