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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.

For an objective material isotropy law, permutation of the principal stretches permutes the labelled principal stresses. This follows by conjugating the left stretch tensor with signed permutation rotation matrices. Each principal stress is symmetric in the other two stretches, but need not be symmetric in all three. The law $\sigma =V^2$ provides the example ${\sigma }_i={\lambda }_i^2$. In an invariant representation $\sigma ={\beta }_{0}I+{\beta }_{1}V+{\beta }_2{V}^2$, the scalar coefficients are symmetric functions of the unordered stretches.

Let $B=A^{\mathsf{T}}A$. Since $\det A>0$, $B$ is a positive-definite matrix. The spectral theorem gives a unique principal square root of a positive semidefinite matrixThen $R^{\mathsf{T}}R=U^{-1}B{U}^{-1}=I$ and $\det R=\det A/\det U=1$, so $R$ is a rotation matrix. Define $V=RU{R}^{\mathsf{T}}$. It is a symmetric matrix, is positive definite, and satisfies $V^2=A{A}^{\mathsf{T}}$. Consequently the polar decomposition in continuum mechanics isHere $U$ and $V$ are the right stretch tensor and left stretch tensor.

For the Cauchy stress tensor, use both material isotropy and material frame indifference. Their respective transformation laws areThe first law and $A=VR$ give $\sigma (A)=\sigma (V)$. Combining the two laws gives $\sigma (QV{Q}^{\mathsf{T}})=Q\sigma (V)Q^{\mathsf{T}}$. In a principal stretch basis, $V=\mathrm{diag}({\lambda }_1,{\lambda }_2,{\lambda }_3)$. Each half-turn about a coordinate axis leaves $V$ unchanged; its conjugation changes the signs of the corresponding off-diagonal stress tensor entries. Those entries must therefore vanish. If two principal stretches coincide, rotations within their eigenspace additionally force the stress tensor to be scalar on that eigenspace. Thus the Cauchy stress tensor and left stretch tensor have common principal axes, including at repeated principal stretches. This proves coaxiality of isotropic elastic stress.

The precise symmetry conclusion is equivariance of isotropic principal stresses. Every permutation of the principal stretches can be realized by a signed permutation rotation matrix, sowhen the indexing is permuted consistently. Equivalently, permuting the arguments permutes the labelled principal stresses in the same way. In particular, ${\sigma }_1$ is symmetric in ${\lambda }_2,{\lambda }_3$, and analogous statements hold for ${\sigma }_2,{\sigma }_3$.

For distinct principal stretches, interpolate the three values ${\sigma }_i$ at the three points ${\lambda }_i$:Uniqueness of this interpolation implies that the scalar coefficients ${\beta }_j$ are symmetric functions of the unordered principal stretches; they may be expressed through the elementary symmetric invariants. Smooth isotropic constitutive laws have the corresponding invariant representation across repeated stretches as well. The wording about symmetric functions must be understood in this collective sense: each labelled principal stress need not be symmetric in all three arguments. For example, the objective isotropic law $\sigma =V^2$ has ${\sigma }_i={\lambda }_i^2$, which disproves that stronger interpretation.

For a real matrix $A$ with positive determinant, the polar decomposition of an invertible real matrix gives unique factorizationswhere $U$ and $V$ are the right stretch tensor and left stretch tensor, respectively. Positivity here means positive definiteness, not merely nonnegative entries.

To prove existence, $C=A^{T}A$ is symmetric and $v^{T}Cv=|Av{|}^2>0$ for every nonzero $v$, because $A$ is invertible. The spectral theorem for real symmetric matrices supplies an orthogonal matrix $Q$ and positive eigenvalues $c_j$ such that $C=Q\mathrm{diag}(c_1,c_2,c_3)Q^T$. Define $U$ using the principal square root of a positive semidefinite matrix construction:Then $R^{T}R=U^{-1}{A}^{T}A{U}^{-1}=I$, so $R$ is orthogonal. Also $\det R=\det A/\det U>0$; an orthogonal matrix has determinant plus or minus one, hence $\det R=1$. Set $V=RU{R}^T$. It is symmetric positive definite, $V^2=A{A}^T$, and $VR=RU=A$.

For uniqueness, any second factorization $A=R_1{U}_1$ with the same properties gives $U_1^2=A^{T}A=C$. Since $U_1$ commutes with its square, it preserves every eigenspace of $C$. On a $c$-eigenspace its own positive eigenvalues must square to $c$, so it equals $\sqrt{c}$ times the identity there. Thus $U_1=U$, then $R_1=A{U}^{-1}=R$, and finally $V$ is fixed too. This proves both uniqueness and the polar decomposition in continuum mechanics, with the same proper rotation in its left and right forms.

For the polar decomposition of an invertible real matrix, start with $C=F^{T}F$. For nonzero $y$, $y^{T}Cy=|Fy{|}^2>0$, since $\det F>0$ makes $F$ invertible. The spectral theorem for real symmetric matrices gives an orthonormal basis $n_a$ with $C{n}_a={\lambda }_a^2{n}_a$ and ${\lambda }_a>0$. DefineThen $U$ is a positive-definite symmetric matrix, $U^2=C$, andbecause $\det U=\sqrt{\det (F^{T}F)}=\det F$. Thus $R$ is an orthogonal matrix with positive determinant. Set $V=RU{R}^T$; it too is a positive-definite symmetric matrix, and $VR=RU=F$. Moreover $V^2=F{F}^T$. This provesThe positive square root is unique: any positive symmetric square root of $C$ commutes with $C$, preserves its eigenspaces and has the positive scalar square root on each of them. Hence $U$, then $R$ and $V$, are unique. These are the right stretch tensor, rotation and left stretch tensor of the polar decomposition in continuum mechanics.

A standard spectral strain measure isTake $f\in C^1(0,\infty )$, $f(1)=0$ and $f^{\prime }(1)=1$, with $f$ strictly increasing; the customary stronger condition $f^{\prime }(\lambda )>0$ gives a regular one-to-one parameterization of all positive principal stretches. The zero condition makes a rigid rotation strain free, monotonicity distinguishes extension from contraction and prevents nonunit stretch from being strain free, and the derivative normalization gives the ordinary infinitesimal strain. Indeed, for $F=I+H$,Under a superposed spatial rotation $F\mapsto QF$, $F^{T}F$ and hence $E^f$ remain unchanged, giving material frame indifference. Choosing $f(\lambda )=\lambda -1$ satisfies all these conditions and gives the Biot strain tensor

The meaning of work-conjugate stress and strain here is that a symmetric material tensor $T^f$ satisfiesfor every admissible rate, with both powers measured per reference volume. To keep the nominal-stress index convention explicit, write $M_{iI}=P_{Ii}$, so $M=P^T$ and the power is $M:\dot{F}$. Differentiating $F=RU$ gives $\dot{F}=R(\Omega U+\dot{U})$, where $\Omega =R^T\dot{R}$ is skew. Conservation of angular momentum gives symmetry of $M{F}^T$, or equivalently of $R^{T}MU$. ThereforeThe first term vanishes because a symmetric tensor has zero contraction with a skew tensor. Since ${\dot{E}}^{(1)}=\dot{U}$, the symmetric Biot stress isIf the nominal stress is instead stored spatial-index first as $M$, the same answer is $\mathrm{sym}(R^{T}M)$. The symmetrization is essential: $PR$ need not itself be symmetric.

Use both material isotropy and material frame indifference. The polar decomposition in continuum mechanics writes $F=VR$, where $V$ is the left stretch tensor. Isotropy removes the reference rotation: $\sigma (F)=\sigma (V)$. Choose an orthonormal eigenbasis of $V$. A half-turn about any principal axis leaves $V$ fixed but reverses appropriate off-diagonal entries of $\sigma (V)$. Frame indifference therefore makes those entries zero. If a principal stretch is repeated, rotations in its eigenspace additionally force the stress to be isotropic there. Thus the principal stresses share the stretch axes: this proves coaxiality of isotropic elastic stress with $V$ and with $B=V^2=F{F}^T$.

For simple shear, the left Cauchy-Green deformation tensor isIts principal axes make an angle $\vartheta$ with the coordinate axes satisfying $2\cot (2\vartheta )=\gamma$ when $\gamma \ne 0$. If the two principal stresses are $s_1,s_2$, rotating their diagonal matrix back givesConsequently the universal simple-shear normal-stress identity isEquivalently, it follows immediately from the off-diagonal entry of $\sigma B-B\sigma =0$. At $\gamma =0$, the undeformed material has isotropic stress, so the identity holds there too. No particular elastic constitutive equation has been used.

For an orientation-preserving invertible deformation gradient, the polar decomposition of an invertible real matrix separates a rotation $R$ from the right stretch tensor $U$ or the left stretch tensor $V$. The rotation is shared by the two factorizations, with $V=RU{R}^T$.