= Solution
Let $\Psi=k_\alpha x^\alpha$. For a nonzero symmetric amplitude, substituting the <plane wave> into the wave equation gives $-(k_\sigma k^\sigma)\epsilon^{\mu\nu}\cos\Psi=0$. The gauge condition gives $-k_\sigma\epsilon^{\nu\sigma}\sin\Psi=0$. Thus a propagating mode satisfies
$$
\boxed{k_\sigma k^\sigma=0,\qquad
k_\sigma\epsilon^{\nu\sigma}=0,\qquad
\epsilon^{\mu\nu}=\epsilon^{\nu\mu}.}
$$
The <wavevector> is null, so the <gravitational wave> propagates at the vacuum <speed of light>. The amplitude is transverse as a four-dimensional tensor in <Lorenz gauge in linearized gravity>. This condition alone does not set all time components or the <trace> to zero: <residual gauge symmetry of linearized gravity> remains. Of the ten symmetric components, four Lorenz constraints and four independent residual coordinate choices leave two physical <gravitational wave polarizations>.
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