 Solution

For a Cartesian second-rank tensor, write $S_{ij}=(A_{ij}+A_{ji})/2$ and $W_{ij}=(A_{ij}-A_{ji})/2$. Under an orthogonal change of frame,where the last two equalities follow by interchanging the dummy indices in $A_{ji}^{\prime }$. Thus the symmetric and antisymmetric parts obey the same tensor transformation law. If $A=S+W$ with $S^T=S$, $W^T=-W$, adding or subtracting its transpose forces $S=(A+A^T)/2$ and $W=(A-A^T)/2$. This also proves uniqueness.

For the given matrix, the scalar part is $a=\mathrm{Tr}(A)/3=3$. The symmetric part is $S=\left(\begin{array}{ccc}1& 3& 2\\ 3& 5& 4\\ 2& 4& 3\end{array}\right)$, so the traceless second-rank tensor is $B=S-3I$. The skew part is $W=\left(\begin{array}{ccc}0& -1& 1\\ 1& 0& 2\\ -1& -2& 0\end{array}\right)$. Comparing with the cross product matrix $[p{]}_{\times }=\left(\begin{array}{ccc}0& -p_3& p_2\\ p_3& 0& -p_1\\ -p_2& p_1& 0\end{array}\right)$ givesThen $A=aI+[p{]}_{\times }+B$ proves the decomposition for every vector. The vector dual to the antisymmetric tensor is an axial vector under reflections, and an ordinary vector under proper rotations.