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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)

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  • $\begingroup$ You might want to clarify your question. To me this seems to be just a change of summation index (from $j$ to $i$ in the first term). $\endgroup$ – Klaus Oct 9 at 16:39
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$$=\sum_i^n a_{ki} x_i + \sum_j^n a_{jk} x_j$$

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....

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

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

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

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

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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.

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  • $\begingroup$ sorry, not a typo in my question... though its a good theory... $\endgroup$ – Leaky Capacitor Oct 9 at 19:39
  • $\begingroup$ I didn't think you had a typo. This answer is not just a theory, it;s the same as yours. There is no difference between $a{jk}x_j$ and $a{ik}x_i$ when you sum over the repeated index. That index is just a dummy variable. Glad you figured it out. Please accept your answer to take the question off the unanswered queue. $\endgroup$ – Ethan Bolker Oct 9 at 20:55
  • $\begingroup$ maybe you could show the detailed steps of how you arrive at that conclusion... even if you change the indices of the second term from i to j. you still have a problem with the indices on being switched on "a" between terms.... and given that k is a constant..if you write out the expansion.... $\endgroup$ – Leaky Capacitor Oct 10 at 18:00
  • $\begingroup$ @LeakyCapacitor See my edit. $\endgroup$ – Ethan Bolker Oct 10 at 21:09

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