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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 $0<\theta <\pi /2$ and $\lambda >\sin 2\theta$, the discriminant is strictly positive. The two roots have positive sum $\lambda /c^2$ and positive product $s^2/c^2$, so both roots are positive and distinct. Each gives exactly one positive $\alpha$ and hence a different right stretch tensor. If $\lambda \ne 1$, neither tensor is the undeformed identity because its third diagonal entry is $\lambda$. Thus two distinct deformed configurations are possible. Excluding $\lambda =1$ removes the case in which one of the two configurations is the undeformed body; it is not needed just for positivity of the roots.

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.

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 $V=RU{R}^T$.

For a right stretch tensor $U={\sum }_a{\lambda }_a{n}_a\otimes n_a$, define $E^f=f(U)={\sum }_{a}f({\lambda }_a)n_a\otimes n_a$. A standard normalized choice has $f\in C^1(0,\infty )$, $f(1)=0$, $f^{\prime }(1)=1$ and $f$ strictly increasing; requiring $f^{\prime }>0$ gives a nonsingular stretch-to-strain parameterization. This strain vanishes exactly for a rigid rotation, is independent of superposed spatial rotations, and agrees to first order with the infinitesimal symmetric displacement gradient. Indeed $F=I+H$ gives $U=I+(H+H^T)/2+O(\Vert H{\Vert }^2)$, and Taylor expansion of $f$ proves the last statement.

For two distinct reference fibre directions $l_{\pm }=(\cos \theta ,\pm \sin \theta ,0)$, with $0<\theta <\pi /2$, consider the volume-preserving deformation with right stretch tensor $U=\mathrm{diag}({\lambda }^{-1/2}\alpha ,{\lambda }^{-1/2}{\alpha }^{-1},\lambda )$, with positive $\lambda ,\alpha$. The inextensible fibre constraint gives ${\alpha }^2{\cos }^2\theta +{\alpha }^{-2}{\sin }^2\theta =\lambda$. Setting $b={\alpha }^2$ gives a quadratic with the displayed roots. There are two distinct positive branches for $\lambda >\sin 2\theta$. The smallest possible axial stretch is $\lambda =\sin 2\theta$, at which ${\alpha }^2=\tan \theta$ and $(U{l}_{+})\cdot (U{l}_{-})=0$. Since both image fibres have unit length, their mutual area cannot exceed one; their area is $\sin 2\theta /\lambda$, which gives the same lower bound. Coincident or antiparallel fibres at the angle endpoints do not give the two-family orthogonality conclusion.