= Solution
Take the <trace> first: the <antisymmetric tensor> has zero diagonal and $X$ is traceless, so $T_{ii}=3t$. Taking symmetric and antisymmetric parts then gives the <scalar, symmetric-traceless and axial tensor decomposition>:
$$
\boxed{t=\frac13T_{kk},\quad A_{ij}=\frac12(T_{ij}-T_{ji}),\quad X_{ij}=\frac12(T_{ij}+T_{ji})-\frac13T_{kk}\delta_{ij}.}
$$
The <Kronecker delta> satisfies $\delta_{ii}=3$, which verifies $X_{ii}=0$. Substitution verifies the original decomposition and also proves its <uniqueness>.
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