= Solution
The <scalar, symmetric-traceless and axial tensor decomposition> fixes the scalar part by the trace:
$$
\boxed{P=\frac13\operatorname{tr}F=\frac43r^2.}
$$
Subtracting this isotropic part from the <symmetric tensor> in part (c) gives
$$
\boxed{\widehat S=\frac13\begin{pmatrix}-x_1^2-4x_2^2&3x_1x_2&0\\3x_1x_2&-4x_1^2-x_2^2&0\\0&0&5r^2\end{pmatrix}.}
$$
Its trace is zero. The <antisymmetric tensor> part is
$$
\boxed{\widehat A=\frac12(F-F^T)=\begin{pmatrix}0&-r^2&x_1-x_2\\r^2&0&-x_1-x_2\\-x_1+x_2&x_1+x_2&0\end{pmatrix}.}
$$
Adding $PI+\widehat S+\widehat A$ reconstructs every entry of $F$.
Subtracting a scalar multiple of the identity preserves every <eigenvector>: if $Sv=\lambda v$, then $(S-PI)v=(\lambda-P)v$. Thus the <principal axes> are \b[the same as in part (c)], with shifted principal values
$$
\boxed{-r^2/3\text{ along }e_r,\qquad-4r^2/3\text{ along }e_\theta,\qquad5r^2/3\text{ along }e_3.}
$$
The two tensors are <coaxial symmetric tensors>. At $r=0$ both vanish, so neither specifies unique axes.
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