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Use linearization around Minkowski spacetime after constant rescaling of $x,y$: set $A=1+\epsilon a(u)$, $B=1+\epsilon b(u)$, $u=t-z$. To first order,Hence $s=a+b$ is affine in $u$, while $h=a-b$ is the freely propagating profile. The affine transverse trace in a diagonal linearized plane wave is pure gauge: it has zero linearized curvature and is locally a coordinate artifact. To see this without imposing extra boundary conditions, use a first-order linearized coordinate gauge transformation $h_{\mu \nu }\mapsto h_{\mu \nu }-{\partial }_{\mu }{\xi }_{\nu }-{\partial }_{\nu }{\xi }_{\mu }$, withThe $tx,ty,zx,zy$ changes cancel and the remaining time/longitudinal changes are proportional to $s^{\prime \prime }=0$. The transverse perturbation becomeswith all time and longitudinal components zero. It is a transverse-traceless gauge plane gravitational wave travelling in the positive $z$-direction, with plus polarization and no cross polarization in these transverse axes.

For freely falling nearby particles, geodesic deviation gives opposite transverse tidal accelerations, since $R_{txtx}=-\epsilon a^{\prime \prime }$ and $R_{tyty}=-\epsilon b^{\prime \prime }=+\epsilon a^{\prime \prime }$ in vacuum. Thus the transverse geodesic deviation eigenvalues have opposite signs; the stretching and compressing roles interchange when the curvature profile changes sign. If $h^{\prime \prime }=0$ as well, there is no linearized tidal wave: that special profile is also locally flat to this order. The exact diagonal metric describes a fixed transverse polarization through its opposite vacuum curvature eigenvalues; the linearized calculation identifies it as plus relative to the chosen axes.