# Tensor identity - Help

1. Jul 31, 2004

### jvicens

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?

2. Jul 31, 2004

### Lonewolf

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.

Last edited: Jul 31, 2004
3. Jul 31, 2004

### robphy

\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*}

4. Jul 31, 2004

### jvicens

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

Last edited: Jul 31, 2004