= Solution
By the <spectral theorem> a <symmetric positive-definite matrix> $C$ has $C=Q\operatorname{diag}(c_i)Q^{\mathsf T}$, with $Q$ <orthogonal> and $c_i>0$. Define $S=Q\operatorname{diag}(\sqrt{c_i})Q^{\mathsf T}$. It is <symmetric> <positive-definite> and $S^2=C$. For <uniqueness>, any <symmetric> <positive-definite> square root $T$ commutes with $C=T^2$, hence preserves each <eigenspace> of $C$. On its $c$-<eigenspace>, $T$ is <symmetric> <positive-definite> and all its <eigenvalues> have square $c$, so all equal $\sqrt c$. Therefore $T=S$, including repeated <eigenspaces>.
For $\det F>0$, put
$$
\boxed{U=(F^{\mathsf T}F)^{1/2},\quad R=FU^{-1},\quad V=RUR^{\mathsf T}=(FF^{\mathsf T})^{1/2}.}
$$
Then $R^{\mathsf T}R=I$ and $\det R=\det F/\det U>0$, so $R\in SO(3)$; also $F=RU=VR$. Any other <polar decomposition in continuum mechanics>[right polar decomposition] has $F^{\mathsf T}F=U^2$, so <uniqueness> of the positive root fixes $U$ and then $R$. The left factor is consequently fixed too.
The principal stretches $v_i$ are the positive <eigenvalues> of $U$, equivalently the <singular values> of $F$. For a <unit vector> $N=\sum_i c_iN_i$ in an orthonormal <eigenbasis> of $U$,
$$
|FN|^2=N^{\mathsf T}U^2N=\sum_i v_i^2c_i^2\leq v_{\max}^2\sum_i c_i^2=v_{\max}^2.
$$
Thus $\boxed{\max_{|N|=1}|FN|=v_{\max}}$, attained precisely by the <unit vectors> in the <eigenspace> of $U$ belonging to $v_{\max}$. With a simple largest <eigenvalue> these are its two opposite unit <eigenvectors>; degeneracy allows every unit combination within that <eigenspace>.
For the <simple shear>,
$$
F=\begin{pmatrix}1&\gamma&0\\0&1&0\\0&0&1\end{pmatrix},\qquad C=F^{\mathsf T}F=\begin{pmatrix}1&\gamma&0\\\gamma&1+\gamma^2&0\\0&0&1\end{pmatrix}.
$$
A line with nonzero direction $N$ keeps its length exactly when
$$
|FN|^2-|N|^2=\gamma N_2(2N_1+\gamma N_2)=0.
$$
Hence $\boxed{N_2=0\quad\text{or}\quad2N_1+\gamma N_2=0}$, with $N_3$ arbitrary. Every segment in the cube <linear dependence>[parallel] to one of these directions preserves length; there is no restriction on its starting point other than remaining in the cube.
The two in-<plane> stretches are $v_\pm=(\sqrt{4+\gamma^2}\pm\gamma)/2$, and the third is one. A maximizing direction for $\gamma>0$ is $N\parallel(1,v_+,0)$, since the $C$ <eigenvalue> equation gives $N_2/N_1=(v_+^2-1)/\gamma=v_+$. Therefore $\boxed{N\to\pm(1,1,0)/\sqrt2}$ as $\gamma\downarrow0$. At exactly zero shear all directions tie; the limit refers to the selected maximizing direction for positive shear. These are the <length-preserving directions and principal directions of simple shear>.
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