= Solution
For a real matrix $A$ with positive <determinant>, the <polar decomposition of an invertible real matrix> gives unique factorizations
$$
\boxed{A=RU=VR,\qquad R\in SO(3),\quad U=U^T>0,\quad V=V^T>0,}
$$
where $U$ and $V$ are the <right stretch tensor> and <left stretch tensor>, respectively. Positivity here means positive definiteness, not merely nonnegative entries.
To prove existence, $C=A^TA$ is symmetric and $v^TCv=|Av|^2>0$ 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 $C=Q\operatorname{diag}(c_1,c_2,c_3)Q^T$. Define $U$ using the <principal square root of a positive semidefinite matrix> construction:
$$
U=Q\operatorname{diag}(\sqrt{c_1},\sqrt{c_2},\sqrt{c_3})Q^T,
\qquad R=AU^{-1}.
$$
Then $R^TR=U^{-1}A^TAU^{-1}=I$, 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 $V=RUR^T$. It is symmetric positive definite, $V^2=AA^T$, and $VR=RU=A$.
For uniqueness, any second factorization $A=R_1U_1$ with the same properties gives $U_1^2=A^TA=C$. 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 $R_1=AU^{-1}=R$, 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.
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