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The polar decomposition in continuum mechanics and material isotropy remove the reference rotation from an elastic Cauchy stress tensor. Half-turns around principal stretch axes leave the diagonal left stretch tensor fixed, so material frame indifference forces the stress's off-diagonal entries to vanish in those axes. Rotations within repeated-stretch eigenspaces also force isotropic stress there. Consequently stress and the left Cauchy-Green deformation tensor are coaxial symmetric tensors.

The scalar, symmetric-traceless and axial tensor decomposition fixes the scalar part by the trace:Subtracting this isotropic part from the symmetric tensor in part (c) givesIts trace is zero. The antisymmetric tensor part isAdding $PI+\hat{S}+\hat{A}$ 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 the same as in part (c), with shifted principal valuesThe two tensors are coaxial symmetric tensors. At $r=0$ both vanish, so neither specifies unique axes.

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.