= Symmetric tensor decomposition relative to an axis
{title2=$T=\alpha I+\beta AA^T+BA^T+AB^T+C$}
For a real symmetric <Cartesian second-rank tensor> $T$ and a nonzero <vector> $A$, require $B\cdot A=0$ and $C$ symmetric and traceless with $CA=0$. Put $q=A^TA$, $s=A^TTA$, $t=\operatorname{tr}T$, and $P=I-AA^T/q$. The unique decomposition is
$$
\alpha=\frac{t-s/q}{2},\quad\beta=\frac{3s/q-t}{2q},\quad B=\frac{TA}{q}-\frac{sA}{q^2},\quad C=PTP-\alpha P.
$$
The <orthogonal projection> $P$ isolates the plane perpendicular to $A$. Splitting $I=P+AA^T/q$ on both sides of $T$ gives its longitudinal, mixed and transverse blocks, proving reconstruction. The transverse <trace> fixes $\alpha$, the longitudinal component fixes $\beta$, and the mixed block fixes $B$, proving uniqueness. The remaining transverse block $C$ has two degrees of freedom. Together $\alpha,\beta,B,C$ carry $1+1+2+2=6$ components, the number in a symmetric three-by-three <matrix>.
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