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A two-index Cartesian tensor has components transforming as $T_{ij}^{\prime }=R_{ik}{R}_{jl}{T}_{kl}$ under an orthogonal change of orthonormal basis. The tensor order here counts indices, rather than the matrix rank. Its scalar, symmetric-traceless and axial tensor decomposition follows by settingThen $T_{ij}=Y{\delta }_{ij}+{\Omega }_{ij}+S_{ij}$, with $\Omega$ antisymmetric and $S$ symmetric and traceless. There are three independent antisymmetric components and five independent symmetric traceless components: three off-diagonal entries and two independent diagonal ones in the latter case.

For the symmetric tensor relative to a nonzero vector $A$, put $q=A_i{A}_i>0$, $s=A_i{T}_{ij}{A}_j$, and $t=T_{ii}$. The symmetric tensor decomposition relative to an axis givesIn particular $A_i{B}_i=s/q-s/q=0$. Define the orthogonal projection $P_{ij}={\delta }_{ij}-A_i{A}_j/q$ onto the plane perpendicular to $A$, and setThis tensor is symmetric and kills $A$. Its trace is $t-s/q-2\alpha =0$. To verify the full decomposition, set $u=A/\sqrt{q}$ and split $T$ into its longitudinal component $(u^{T}Tu)u{u}^T$, two mixed components, and $PTP$. The mixed terms are exactly $B{A}^T+A{B}^T$, and $\alpha +\beta q=s/q$ gives the longitudinal coefficient. Thus all requested constraints and the reconstruction hold.