= Solution
Insert $D=\operatorname{diag}(1,-1)$ at the first undecorated site. In the <eigenbasis> of the <transfer matrix>, $D$ exchanges $v_+$ and $v_-$, and both diagonal <matrix elements> vanish. Explicitly,
$$
\boxed{\langle\sigma_1\rangle_L=\frac{\operatorname{tr}(DV^L)}{\operatorname{tr}V^L}
=\frac{\lambda_+^L v_+^{\mathsf T}Dv_++\lambda_-^L v_-^{\mathsf T}Dv_-}{\lambda_+^L+\lambda_-^L}=0.}
$$
Thus its large-$L$ limit is zero at every fixed $T>0$. The result is exact even before taking that limit. Equivalently, simultaneous reversal of all <spins> pairs configurations of equal <energy> and opposite $\sigma_1$. Positivity of the <transfer matrix> at fixed <Temperature>[positive temperature] prevents a degenerate dominant <eigenvalue> and <spontaneous magnetization> in this one-dimensional zero-field chain. Selecting one aligned state at strictly <Absolute zero>[zero temperature] is a different order of limits.
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