# Tensor identity - Help

by jvicens
Tags: identity, tensor
 P: 18 I'm having some trouble to prove the following tensor identity shown below in Einstein's summation convention: $$(a_{ij}+a_{ji})x_{i}x_{j}=2a_{ij}x_{i}x_{j}$$ I expanded the terms but when I did group them I didn't get the identity. The only way I could get the identity is if $$a_{ij}=a_{ji}$$ and I don't see a reason why this would be so. Obviously I'm missing something. Can somebody tell what is it that I'm doing wrong?
 P: 333 It's true that $$a_{ij}x_{i}x_{j} = a_{ji}x_{i}x_{j}$$. Putting in the summation signs may help you to see this.
 Sci Advisor HW Helper PF Gold P: 4,097 To elaborate on Lonewolf's reply... \begin{align*} a_{ij}x_{i}x_{j} &= a_{ji}x_{j}x_{i}&&\text{relabel repeated [dummy] indices}\\ &= a_{ji}x_{i}x_{j}&&\text{reorder the writing of the x factors}\\ \end{align*} You'll learn that $\frac{1}{2}(a_{ij} + a_{ji})$ is called "the symmetric part of $a_{ij}$", and is written as $a_{(ij)}$. Hence, your identity can be written $$2a_{(ij)}x_{i}x_{j} =2a_{ij}x_{i}x_{j}$$. Here is an "index gymnastics" proof, starting with half of your right-hand-side: \begin{align*} a_{ij}x_{i}x_{j} &= a_{ij}x_{(i}x_{j)}&&\text{since } x_{i}x_{j} = x_{(i}x_{j)} \\ &= a_{(ij)}x_{i}x_{j}&&\text{since i and j are being symmetrized} \end{align*}
P: 18

## Tensor identity - Help

Guys, thanks to you help I got it right this time after some thinking and some calculations. I realized one of my problems was the fact that when I was looking at
$$a_{ij}=a_{ji}$$
I was thinking of it as when we say
$$a=b$$
The way I should look at that expression is not isolated from the rest of the other terms. The important thing, I think, is not the term itself isolated but how the summation terms comes up after the entire summation is expanded.
I can continue now with the next page of my book

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