 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 $S=v{v}^T+c{r}^2{e}_3{e}_3^T$ has principal axeswith eigenvalues $r^2,0,c{r}^2$, respectively. This follows directly from the outer product: $v{v}^{T}v=r^{2}v$, while it annihilates vectors perpendicular to $v$. The orthogonal matrix with rows $e_r^T,e_{\theta }^T,e_3^T$ gives $MS{M}^T=\mathrm{diag}(r^2,0,c{r}^2)$.

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.