Undergrad Transfer rank2 tensor to a new basis

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The trace of the sigma tensor should remain consistent across both the new and old bases, but discrepancies are noted in the calculations. The discussion highlights the expectation of off-diagonal entries in the new basis, which are absent. A specific tensor expression is provided, indicating a method to express the new basis vectors in terms of the old ones. By substituting these expressions into the tensor equation and applying the distributive law, a new tensor representation can be derived. Clarification on tensor manipulation rules is sought to resolve the inconsistencies.
GGGGc
The trace of the sigma should be the same in both new and old basis. But I get a different one. Really appreciate for the help.
I’ll put the screen shot in the comment part
 
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I have no idea about your calculations since I haven't really learned the rules of manipulations of tensors. However, I am surprised that you don't get off-diagonal entries although the new basis looks as if there should be several of them.

I see that
$$
\sigma_{ij}=\underline{\hat{e}}^{(1)} \otimes \underline{\hat{e}}^{(1)}+\underline{\hat{e}}^{(2)}\otimes \underline{\hat{e}}^{(2)}+2\cdot\underline{\hat{e}}^{(3)}\otimes \underline{\hat{e}}^{(3)}
$$
Hence, I would write ##\underline{\hat{e}}^{(k)}=\alpha_k \underline{\hat{e}}'^{(1)}+\beta_k \underline{\hat{e}}'^{(2)}+\gamma_k \underline{\hat{e}}'^{(3)},## determine the ## \alpha_k\, , \,\beta_k\, , \,\gamma _k,## substitute all of them in the first equation and rearrange everything with the distributive law to obtain an equation
$$
\sigma'_{ij}=\sum_{p,q,r=1}^3 s_{pqr} \cdot \underline{\hat{e}}'^{(p)}\otimes \underline{\hat{e}}'^{(q)}\otimes \underline{\hat{e}}'^{(r)}
$$
 

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