= Solution
A <deformation> maps a material point in the reference body to its current position, $y:\Omega\to\mathbb R^3$, with <deformation gradient> $F=Dy$. A <homogeneous deformation> has constant <gradient> and is affine, $y(X)=FX+b$.
If $\det F>0$, the matrix $F^TF$ is symmetric positive definite. The matrix-square-root theorem says that any such matrix has a unique symmetric positive definite square root, obtained by taking positive square roots in its orthogonal <spectral decomposition>. Set
$$
\boxed{U=(F^TF)^{1/2},\qquad R=FU^{-1}.}
$$
Then $R^TR=U^{-1}F^TFU^{-1}=I$ and $\det R=\det F/\det U>0$, so $R\in\mathrm{SO}(3)$ and $F=RU$. If also $F=\widetilde R\widetilde U$, then $F^TF=\widetilde U^2$; <uniqueness> of the positive square root gives $\widetilde U=U$ and then $\widetilde R=R$. This proves \b[existence and <uniqueness> of <polar decomposition>].
The principal stretches are the positive <eigenvalues> of $U$, equivalently the <singular values> of $F$ or square roots of the <eigenvalues> of $F^TF$.
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