= Solution
The unique symmetric and antisymmetric decomposition is $S=(T+T^T)/2$, $A=(T-T^T)/2$. Both pieces are <Cartesian tensors>: transposing $T'=RTR^T$ gives $(T')^T=RT^TR^T$, so $S'=RSR^T$ and $A'=RAR^T$. Hence this is a decomposition into a <symmetric tensor> and an <antisymmetric second-rank tensor>, rather than just a decomposition of numerical arrays.
For the <axial vector of an antisymmetric second-rank tensor>, contract the two <Levi-Civita symbols>:
$$
\epsilon_{kij}a_k=\frac12\epsilon_{kij}\epsilon_{klm}A_{lm}=\frac12(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})A_{lm}=A_{ij}.
$$
Thus
$$
\boxed{A=\begin{pmatrix}0&a_3&-a_2\\-a_3&0&a_1\\a_2&-a_1&0\end{pmatrix}.}
$$
With this convention $Ax=x\times a$, so the displayed matrix is the negative of the <cross-product matrix> for $a\times x$.
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