Why is principal strain parallel to principal stress for high symmetry crystals

Question

Let say we obtain $\epsilon_{11} = S_{1111}\sigma_{11}$, for a single stress in a 11 direction, meaning that $E=\dfrac{1}{S_{1111}}$.
Due to high isotropy of material the Poisson ratio is constant. This means that $\epsilon_{2}=S_{2211}\sigma_1 = S_{1122}\sigma_1$
$\epsilon_{2}=-v\epsilon_1$
hence $-v\epsilon_{1}=S_{1122}\sigma_1$

$\epsilon_1=\dfrac{1}{E} (\sigma_1-v(\sigma_2+\sigma_3))$
$\epsilon_2=\dfrac{1}{E} (\sigma_2-v(\sigma_1+\sigma_3))$
$\epsilon_3=\dfrac{1}{E} (\sigma_3-v(\sigma_1+\sigma_2))$

And then its written, that since the elastic response is so symmetrically constrained, the principal strains are parallel to principal stresses

Answer 1

The alternative (i.e., that the principal strains are angled relative to the principal stresses) is not tenable because if the material is isotropic, then Nature has no way of knowing how that angle should be oriented.

Put another way, if you uniaxially load an axisymmetric isotropic sample under tension, then there's no single direction other than that axis that can exhibit tensile strain and also remain unchanged as I spin the sample around the axis (which would leave an isotropic sample looking unchanged).