= Solution
For a nonzero plane-wave <trace-reversed metric perturbation>,
$$
\Box\left(A^{\alpha\beta}e^{ik_\gamma x^\gamma}\right)
=-(k_\gamma k^\gamma)A^{\alpha\beta}e^{ik_\gamma x^\gamma}.
$$
The <wave equation> therefore requires a null <wave vector>. The <Lorenz gauge in linearized gravity> supplies transversality:
$$
\boxed{k_\gamma k^\gamma=0,\qquad k_\alpha A^{\alpha\beta}=0.}
$$
The amplitude is symmetric because it represents a <metric perturbation>; its real part is the physical wave. The two displayed conditions do not by themselves make it spatially transverse and traceless: the additional reduction to <transverse-traceless gauge> uses the remaining <residual gauge symmetry of linearized gravity>.
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