= Principal axes of a planar rank-one tensor block
For $v=(a,b,0)$ with $r=\sqrt{a^2+b^2}>0$, the <symmetric tensor> $S=vv^T+cr^2e_3e_3^T$ has <principal axes>
$$
e_r=\frac1r(a,b,0),\qquad e_\theta=\frac1r(-b,a,0),\qquad e_3=(0,0,1),
$$
with <eigenvalues> $r^2,0,cr^2$, respectively. This follows directly from the <outer product>: $vv^Tv=r^2v$, while it annihilates vectors perpendicular to $v$. The orthogonal matrix with rows $e_r^T,e_\theta^T,e_3^T$ gives $MSM^T=\operatorname{diag}(r^2,0,cr^2)$.
If $a,b$ depend on position, this is a pointwise eigenframe, not necessarily one fixed Cartesian change of axes throughout space. For $v=0$, the displayed tensor vanishes and every orthonormal basis is a principal basis. Subtracting a scalar multiple of the identity leaves its eigenvectors unchanged and shifts every eigenvalue by the same amount; this is a simple case of <coaxial symmetric tensors>.
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