Proving T_{ijk} is a Third Rank Tensor and its Transformation Properties

[latentcorpse]

$T_{ijk}$ is an array with 27 components which is not known to represent a tensor. If for every second rank tensor, $R_{ij}$, the quantity $v_i=T_{ijk}R_{jk}$ is always a vector, show that $T_{ijk}$ is a third rank tensor.

I've managed the bit above. Just stuck on the next part:

If $R_{ij}$ is any symmetric tensor and $v_i$ is again always a vector, what can be said about the transformation properties of $T_{ijk}$ and $T_{ijk}+T_{ikj}$?

 

[Jhero]

Since v_i=T_{ijk}R_{jk} is always a vector, we can say that T_{ijk}R_{jk} is also a vector for any symmetric tensor R_{ij}. This means that T_{ijk} must transform like a third rank tensor, since it is able to produce a vector when contracted with a second rank symmetric tensor.

Now, considering T_{ijk}+T_{ikj}, we can see that the indices i and j are interchangeable due to the symmetry of the tensor R_{ij}. This means that the resulting vector v_i=T_{ijk}R_{jk} will be the same regardless of whether we use T_{ijk} or T_{ikj}. Therefore, we can say that T_{ijk}+T_{ikj} has the same transformation properties as T_{ijk}, and both are third rank tensors.