= Solution
Use <matrix> notation and set $s=\mathbf v^T\mathbf v>0$, $t=\mathbf v^TT\mathbf v$, $\tau=\operatorname{tr}T$. Contracting the proposed decomposition with $\mathbf v$ and taking its <trace>, using the stipulated transverse and trace conditions, gives
$$
T\mathbf v=(A+Bs)\mathbf v+s\mathbf C,\qquad t=sA+s^2B,\qquad\tau=3A+sB.
$$
These relations determine both scalars and the transverse <vector> uniquely:
$$
\boxed{A=\frac12\left(\tau-\frac ts\right),\qquad B=\frac{3t-s\tau}{2s^2},\qquad\mathbf C=\frac{T\mathbf v-(t/s)\mathbf v}{s}.}
$$
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 define
$$
\boxed{D=PTP-AP.}
$$
Both terms are symmetric, $D\mathbf v=0$, and
$$
\operatorname{tr}D=\operatorname{tr}(TP)-A\operatorname{tr}P=\tau-\frac ts-2A=0.
$$
Expand $T=(P+Q)T(P+Q)$ 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 $PTP=AP+D$. Consequently
$$
T=AI+\left(\frac t{s^2}-\frac As\right)\mathbf v\mathbf v^T+\mathbf C\mathbf v^T+\mathbf v\mathbf C^T+D,
$$
and 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 $2\times2$ block. Its zero <trace> leaves two independent entries, of the form $\bigl(\begin{smallmatrix}d&e\\e&-d\end{smallmatrix}\bigr)$. Together with the two independent scalars $A,B$, the total is \b[$1+1+2+2=6$ parameters], exactly $3(3+1)/2$, the <dimension> of real symmetric $3\times3$ matrices. The recovery formulas also prove uniqueness, so this is an actual parameterization without redundant degrees of freedom.
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