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Any tensor of rank 2 can be rewritten as:

$$A_{bc} = \frac{1}{2}(A_{bc} + A_{cb}) + \frac{1}{2}(A_{bc}-A_{cb})$$

I can understand how that works. My question is:

Prove that (independently): $$\frac{1}{2}(A_{bc} + A_{cb})$$ is symmetric, and $$\frac{1}{2}(A_{bc}-A_{cb})$$ is antisymmetric.

Is there a proof, or is this just a definition? Thanks in advance!

(NOTE: I don't want to see how these terms being symmetric and antisymmetric explains the expansion of a tensor. I see that if it is symmetric, the second relation is 0, and if antisymmetric, the first first relation is zero, so that you recover the same tensor)

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    $\begingroup$ Try this: take one of your expressions, exchange b and c, and use the fact that addition is commutative. Does the result match the definition of (anti-)symmetry? $\endgroup$ – Andrew Oct 11 '14 at 14:15
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It's almost the defition. A tensor $T_{ab}$ of rank $2$ is symmetric if, and only if, $T_{ab}=T_{ba}$, and antisymmetric if, and only if, $T_{ab}=-T_{ba}$. So from this definition you can easily check that this decomposition indeed yields a symmetric and antisymmetric part.

Edit: Let $S_{bc}=\dfrac{1}{2}\left(A_{bc}+A_{cb}\right)$. Then $$S_{cb}=\dfrac{1}{2}\left(A_{cb}+A_{bc}\right)=\dfrac{1}{2}\left(A_{bc}+A_{cb}\right)=S_{bc},$$ so, $S_{bc}$ is symmetric. On the same way, if $T_{bc}=\dfrac{1}{2}\left(A_{bc}-A_{cb}\right)$, we have $$T_{cb}=\dfrac{1}{2}\left(A_{cb}-A_{bc}\right)=-\dfrac{1}{2}\left(A_{bc}-A_{cb}\right)=-T_{bc},$$ and $T_{bc}$ is antisymmetric.

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  • $\begingroup$ is it not possible to show that for any $A_{bc}$, $1/2(A_{bc} + A_{cb}$ is symmetric?that's what I want to see.... $\endgroup$ – GRrocks Oct 11 '14 at 14:11
  • $\begingroup$ Edited the answer for clearer explanation. $\endgroup$ – Mateus Sampaio Oct 11 '14 at 14:19

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