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)
Best Answer
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.