= Solution
For a nonzero, nonconstant vacuum <plane gravitational wave in linearized gravity> with real wave covector $k$, the <Linearized Einstein equations> and <Lorenz gauge in linearized gravity> require
$$
\boxed{H_{\mu\nu}=H_{\nu\mu},\qquad k^\mu H_{\mu\nu}=0,\qquad k^\mu k_\mu=0.}
$$
The zero-amplitude solution places no restriction on $k$; a constant mode $k=0$ is not a propagating wave. Under the sign convention for the infinitesimal change specified here, the <trace-reversed metric perturbation> changes by
$$
\delta\bar h_{\mu\nu}=\partial_\mu\xi_\nu+\partial_\nu\xi_\mu-\eta_{\mu\nu}\partial_\rho\xi^\rho,\qquad \partial^\mu\delta\bar h_{\mu\nu}=\Box\xi_\nu.
$$
Thus the <residual gauge symmetry of linearized gravity> is characterized by \b[$\Box\xi_\nu=0$]. The proposed plane-wave form obeys this for every constant complex $X_\nu$, because $k$ is null. Changing the sign used to name $\xi$ would reverse both gauge formulas, with no physical consequence.
Rotate the spatial axes so the wave propagates along $+z$, choosing $k_\mu=(-\omega,0,0,\omega)$ with $\omega>0$. The <Lorenz gauge in linearized gravity> says $H_{3\nu}=-H_{0\nu}$; in particular $H_{03}=-H_{00}$, $H_{33}=H_{00}$ and $H_{3i}=-H_{0i}$ for $i=1,2$. The amplitude change is
$$
\delta H_{\mu\nu}=i\bigl(k_\mu X_\nu+k_\nu X_\mu-\eta_{\mu\nu}k^\rho X_\rho\bigr).
$$
Choose
$$
X_1=\frac{H_{01}}{i\omega},\qquad X_2=\frac{H_{02}}{i\omega},\qquad X_3-X_0=\frac{iH_{00}}{\omega},\qquad X_0+X_3=\frac{H_{11}+H_{22}}{2i\omega}.
$$
Indeed, $\delta H_{00}=i\omega(X_3-X_0)$ and $\delta H_{0i}=-i\omega X_i$ cancel the time components, while $\delta H_{11}=\delta H_{22}=-i\omega(X_0+X_3)$ cancels the transverse trace. The preserved <Lorenz gauge in linearized gravity> then cancels all longitudinal components. This is an <explicit plane-wave reduction to transverse-traceless gauge>, yielding
$$
\boxed{H'_{\mu\nu}=\begin{pmatrix}0&0&0&0\\0&H_+&H_\times&0\\0&H_\times&-H_+&0\\0&0&0&0\end{pmatrix},\qquad H_+=\frac{H_{11}-H_{22}}2,\quad H_\times=H_{12}.}
$$
The <trace-reversed metric perturbation> has zero trace in this <transverse-traceless gauge>, so $h'=\bar h'$. The null dispersion relation is $\omega^2=|\mathbf k|^2$, giving the speed of light. Only components perpendicular to the direction of propagation remain, and the two independent amplitudes give the plus and cross <gravitational wave polarizations>.
These are physical transverse tidal distortions, rather than just a convenient display of the <metric perturbation>: in <transverse-traceless gauge>, $R^{(1)}_{0i0j}=-\tfrac12\partial_t^2h'_{ij}$, so a freely falling detector has $\ddot S^i=\tfrac12\ddot h'_{ij}S^j$ at first order. There is no longitudinal tidal acceleration. The plus polarization stretches one transverse axis while compressing the other; the cross polarization does the same along axes rotated by $\pi/4$.
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