= Solution
For the <polar decomposition of an invertible real matrix>, start with $C=F^TF$. For nonzero $y$, $y^TCy=|Fy|^2>0$, since $\det F>0$ makes $F$ invertible. The <spectral theorem for real symmetric matrices> gives an <orthonormal basis> $n_a$ with $Cn_a=\lambda_a^2n_a$ and $\lambda_a>0$. Define
$$
U=\sum_a\lambda_a n_a\otimes n_a,\qquad R=FU^{-1}.
$$
Then $U$ is a <positive-definite symmetric matrix>, $U^2=C$, and
$$
R^TR=U^{-1}F^TFU^{-1}=I,\qquad
\det R=\frac{\det F}{\det U}=1,
$$
because $\det U=\sqrt{\det(F^TF)}=\det F$. Thus $R$ is an <orthogonal matrix> with positive determinant. Set $V=RUR^T$; it too is a <positive-definite symmetric matrix>, and $VR=RU=F$. Moreover $V^2=FF^T$. This proves
$$
\boxed{F=RU=VR,\quad U=(F^TF)^{1/2},\quad V=(FF^T)^{1/2},\quad R^TR=I,\quad\det R=1.}
$$
The 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> is
$$
E^f=f(U)=\sum_a f(\lambda_a)n_a\otimes n_a.
$$
Take $f\in C^1(0,\infty)$, $f(1)=0$ and $f'(1)=1$, with $f$ strictly increasing; the customary stronger condition $f'(\lambda)>0$ 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 $F=I+H$,
$$
U=I+\frac{H+H^T}{2}+O(\|H\|^2),\qquad
E^f=\frac{H+H^T}{2}+o(\|H\|).
$$
Under a superposed spatial rotation $F\mapsto QF$, $F^TF$ 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>
$$
\boxed{E^{(1)}=U-I.}
$$
The meaning of <work-conjugate stress and strain> here is that a symmetric material <tensor> $T^f$ satisfies
$$
T^f:\dot E^f=P_{Ii}\dot F_{iI}
$$
for 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 $MF^T$, or equivalently of $R^TMU$. Therefore
$$
\begin{aligned}
M:\dot F
&=\operatorname{tr}((R^TM)^T\Omega U)
+(R^TM):\dot U\\
&=\operatorname{tr}((R^TMU)^T\Omega)
+\operatorname{sym}(R^TM):\dot U\\
&=\operatorname{sym}(R^TM):\dot U.
\end{aligned}
$$
The first term vanishes because a symmetric <tensor> has zero contraction with a skew <tensor>. Since $\dot E^{(1)}=\dot U$, the <symmetric Biot stress> is
$$
\boxed{T^{(1)}=\operatorname{sym}(PR)
=\frac12(PR+R^TP^T).}
$$
If the <nominal stress> is instead stored spatial-index first as $M$, the same answer is $\operatorname{sym}(R^TM)$. The symmetrization is essential: $PR$ need not itself be symmetric.
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