 Solution

Let $s=A_i{A}_i>0$, $q=A_i{T}_{ij}{A}_j$ and $\tau =T_{ii}$. Taking the trace of the proposed decomposition givessince $A_i{B}_i=0$ and $C_{ii}=0$. Contracting twice with $A$ gives $q=\alpha s+\beta s^2$. Solving these two scalar equations yieldsA single contraction gives $T_{ij}{A}_j=(\alpha +\beta s)A_i+s{B}_i$, and $\alpha +\beta s=q/s$. HenceFinally setThis is manifestly symmetric. The formulas give $A_i{B}_i=q/s-q/s=0$; substituting them into $C_{ij}{A}_j$ gives zero, and taking the trace gives $C_{ii}=0$. Since $C$ is symmetric, $A_i{C}_{ij}=0$ also holds.

A geometric form makes the axis decomposition of a symmetric matrix explicit. With the transverse orthogonal projection $P_{ij}={\delta }_{ij}-A_i{A}_j/s$,Here $\alpha$ is half the transverse trace and $C$ is its symmetric traceless rank-two tensor part. Every term was forced by contractions, so the decomposition is unique.