= Solution
A two-index <Cartesian tensor> has components transforming as $T'_{ij}=R_{ik}R_{jl}T_{kl}$ 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 setting
$$
Y=\frac{T_{kk}}3,\qquad\Omega_{ij}=\frac{T_{ij}-T_{ji}}2,\qquad S_{ij}=\frac{T_{ij}+T_{ji}}2-Y\delta_{ij}.
$$
Then $T_{ij}=Y\delta_{ij}+\Omega_{ij}+S_{ij}$, with $\Omega$ antisymmetric and $S$ symmetric and traceless. There are \b[three independent antisymmetric components] and \b[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 $q=A_iA_i>0$, $s=A_iT_{ij}A_j$, and $t=T_{ii}$. The <symmetric tensor decomposition relative to an axis> gives
$$
\boxed{\alpha=\frac12\left(t-\frac sq\right),\qquad\beta=\frac{3s/q-t}{2q},\qquad B_i=\frac{T_{ij}A_j}{q}-\frac{sA_i}{q^2}.}
$$
In particular $A_iB_i=s/q-s/q=0$. Define the <orthogonal projection> $P_{ij}=\delta_{ij}-A_iA_j/q$ onto the plane perpendicular to $A$, and set
$$
\boxed{C_{ij}=P_{ik}T_{kl}P_{lj}-\alpha P_{ij}.}
$$
This <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 $(u^TTu)uu^T$, two mixed components, and $PTP$. The mixed terms are exactly $BA^T+AB^T$, and $\alpha+\beta q=s/q$ gives the longitudinal coefficient. Thus all requested constraints and the reconstruction hold.
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