= Solution
Write $A_{ij}=\partial_j u_i$. The trace fixes the isotropic part, and splitting the remaining matrix into its symmetric and antisymmetric parts gives the <scalar, symmetric-traceless and axial tensor decomposition>:
$$
\boxed{p=\frac13\partial_i u_i,\qquad s_{ij}=\frac12(\partial_j u_i+\partial_i u_j)-\frac13(\partial_k u_k)\delta_{ij}.}
$$
Here $s_{ij}=s_{ji}$ and $s_{ii}=0$. The antisymmetric part is $a_{ij}=(\partial_j u_i-\partial_i u_j)/2$. Its associated axial vector is
$$
\boxed{\omega_k=\frac12\epsilon_{kij}\partial_j u_i=-\frac12(\nabla\times u)_k.}
$$
Indeed the contraction identity $\epsilon_{ijk}\epsilon_{k\ell m}=\delta_{i\ell}\delta_{jm}-\delta_{im}\delta_{j\ell}$ gives $\epsilon_{ijk}\omega_k=(\partial_j u_i-\partial_i u_j)/2$. Adding the three parts reconstructs $A_{ij}$.
\b[The minus sign is required by the index order] $A_{ij}=\partial_j u_i$. The <symmetric gradient and curl decomposition> written instead for $\partial_i u_j$ has the opposite axial sign. The <divergence> supplies $3p$, the trace-free symmetric part describes shape change, and the antisymmetric part describes local rotation.
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