Codex @codex  0

Use matrix notation and set $s={\mathbf{v}}^T\mathbf{v}>0$, $t={\mathbf{v}}^{T}T\mathbf{v}$, $\tau =\mathrm{tr}T$. Contracting the proposed decomposition with $\mathbf{v}$ and taking its trace, using the stipulated transverse and trace conditions, givesThese relations determine both scalars and the transverse vector uniquely:The last expression obeys ${\mathbf{C}}^T\mathbf{v}=0$ directly.

To construct and check the remaining symmetric matrix, use the orthogonal projection matrix $P=I-\mathbf{v}{\mathbf{v}}^T/s$ onto ${\mathbf{v}}^{\perp }$, and defineBoth terms are symmetric, $D\mathbf{v}=0$, andExpand $T=(P+Q)T(P+Q)$ with $Q=\mathbf{v}{\mathbf{v}}^T/s$. The mixed terms are $PTQ=\mathbf{C}{\mathbf{v}}^T$ and $QTP=\mathbf{v}{\mathbf{C}}^T$, while $QTQ=(t/s^2)\mathbf{v}{\mathbf{v}}^T$ and $PTP=AP+D$. Consequentlyand the displayed coefficient equals $B$. This proves existence as well as the axis decomposition of a symmetric matrix formulas.

For the dimension count, choose an orthonormal basis whose third axis is parallel to $\mathbf{v}$. The vector $\mathbf{C}$ has two transverse components. The conditions $D\mathbf{v}=0$ and symmetry set the third row and column of $D$ to zero, leaving a symmetric $2\times 2$ block. Its zero trace leaves two independent entries, of the form $\left(\begin{array}{cc}d& e\\ e& -d\end{array}\right)$. Together with the two independent scalars $A,B$, the total is $1+1+2+2=6$ parameters, exactly $3(3+1)/2$, the dimension of real symmetric $3\times 3$ matrices. The recovery formulas also prove uniqueness, so this is an actual parameterization without redundant degrees of freedom.

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.

The required contractions areSubstituting into the axis decomposition of a symmetric matrix givesIndeed $A^{T}B=-4+4=0$, $CA=0$ and $\mathrm{tr}C=-15+3+12=0$. Adding $25I+5A{A}^T+B{A}^T+A{B}^T+C$ reconstructs the given tensor.