Einstein Summation: $a_{kj}x_{j} + a_{ik} x_{i} = (a_{ik} + a_{ki})x_i$

Question

If k is a constant, and $(1\le i \le n)$, and $(1\le j \le n)$.

Why does this Einstein Summation:

$$a_{kj}x_{j} + a_{ik} x_{i}$$

Equal this Einstein Summation: $$a_{kj}x_{j} + a_{ik} x_{i} = (a_{ik} + a_{ki})x_i$$

(From Schaum's Outlines: Tensor Calculus, 2011, page 5-6, problem 1.10)

Answer 1

It's just the distributive law. Write out an example with $n=2$ so only two values $x_1$ and $x_2$ and you will see that.

In the second term the name of the index is $i$ instead of $j$ but that's irrelevant. It could have been written $$ a_{jk}x_j. $$

Edit in response to comment.

$$ \begin{align} \sum_{j=1}^2 a_{kj}x_j + \sum_{i=1}^2 a_{1k }x_i & = (a_{k1}x_1 + a_{k2}x_2) + (a_{1k}x_1 + a_{2k}x_2) \\ &= (a_{k1}x_1 + a_{1k}x_1) + (a_{k2}x_2 + a_{2k}x_2)\\ & = (a_{k1} + a_{1k})x_1 + (a_{k2} + a_{2k})x_2 \\ & = \sum_{z=1}^2 (a_{kz} + a_{zk})x_z \ . \end{align} $$

The $k$ is fixed throughout. It doesn't matter what indices you use for summation. I chose $z$ for the last sum.

Answer 2

$$=\sum_i^n a_{ki} x_i + \sum_j^n a_{jk} x_j$$

At this point we make a change of variables in the second summation by replacing j with i because its equivalent, and by doing so allows us to further simplify the expression....

$$=\sum_i^n a_{ki} x_i + \sum_i^n a_{ik} x_i$$

$$=\sum_i^n \bigg( a_{ki} x_i + a_{ik} x_i \bigg)$$

$$=\sum_i^n x_i\bigg( a_{ki} + a_{ik} \bigg)$$

$$=x_i ( a_{ki} + a_{ik} )$$