= Solution
Let $r=\sqrt{x_1^2+x_2^2}$. Averaging each pair of transposed entries gives the <symmetric tensor>
$$
\boxed{S=\frac12(F+F^T)=\begin{pmatrix}x_1^2&x_1x_2&0\\x_1x_2&x_2^2&0\\0&0&3r^2\end{pmatrix}.}
$$
For $r>0$, the upper block is the <outer product> of $(x_1,x_2)$ with itself. It sends that vector to $r^2$ times itself and annihilates its perpendicular. Thus its orthonormal <principal axes> and corresponding <eigenvalues> are
$$
\boxed{\begin{array}{c|c|l}
\text{axis}&\text{principal value}&\text{geometric direction}\\\hline
e_r=(x_1,x_2,0)/r&r^2&\text{radially outward from the }x_3\text{-axis}\\
e_\theta=(-x_2,x_1,0)/r&0&\text{tangent to a circle about the }x_3\text{-axis}\\
e_3=(0,0,1)&3r^2&\text{parallel to the }x_3\text{-axis}
\end{array}.}
$$
Taking these vectors as the rows of the change-of-components matrix gives
$$
\boxed{M(x)=\begin{pmatrix}x_1/r&x_2/r&0\\-x_2/r&x_1/r&0\\0&0&1\end{pmatrix},\qquad MSM^T=\operatorname{diag}(r^2,0,3r^2).}
$$
This matrix is orthogonal with determinant one. These are the <principal axes of a planar rank-one tensor block>.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-2-2015-ib-principal-axes.png]
{title=The radial and tangential principal directions rotate with position; the third principal axis is perpendicular to this plane}
The matrix is a pointwise frame transformation: because its angle varies with position, it does not represent one fixed Cartesian rotation diagonalizing the entire field everywhere. On the axis $r=0$, the symmetric tensor is zero; all principal values coincide and any orthonormal axes may be chosen. The displayed formula for $M$ is therefore used only off that axis.
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