= Solution
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 matrix>
$$
U=B^{1/2},\qquad R=AU^{-1}.
$$
Then $R^{\mathsf T}R=U^{-1}BU^{-1}=I$ and $\det R=\det A/\det U=1$, so $R$ is a <rotation matrix>. Define $V=RUR^{\mathsf T}$. It is a <symmetric matrix>, is positive definite, and satisfies $V^2=AA^{\mathsf T}$. Consequently the <polar decomposition in continuum mechanics> is
$$
\boxed{A=RU=VR,\quad U=(A^{\mathsf T}A)^{1/2},\quad V=(AA^{\mathsf T})^{1/2}.}
$$
Here $U$ 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 are
$$
\sigma(AQ)=\sigma(A),\qquad \sigma(QA)=Q\sigma(A)Q^{\mathsf T},\qquad Q\in SO(3).
$$
The first law and $A=VR$ give $\sigma(A)=\sigma(V)$. Combining the two laws gives $\sigma(QVQ^{\mathsf T})=Q\sigma(V)Q^{\mathsf T}$. In a <principal stretch> basis, $V=\operatorname{diag}(\lambda_1,\lambda_2,\lambda_3)$. 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 \b[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>, so
$$
\sigma_i(\lambda_{\pi(1)},\lambda_{\pi(2)},\lambda_{\pi(3)})
=\sigma_{\pi(i)}(\lambda_1,\lambda_2,\lambda_3)
$$
when 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 $\lambda_i$:
$$
\sigma=\beta_0I+\beta_1V+\beta_2V^2,\qquad
\sigma_i=\beta_0+\beta_1\lambda_i+\beta_2\lambda_i^2.
$$
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.
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