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A spatial rotation and rescaling put the future null vector in the form $k^{\mu }=(1,0,0,1)$, so the profile depends on $z-t$. Transversality gives four linear relations among the ten symmetric components. The four residual gauge functions satisfying $\square {\xi }^{\mu }=0$ remove the time and longitudinal components; the remaining trace can be removed by the residual transformation indicated in the question. The resulting transverse-traceless gauge isThe two arbitrary functions are the plus and cross gravitational-wave polarizations. They are the two physical degrees of freedom left after the four gauge conditions and four residual coordinate freedoms are removed.

All time components of the perturbation vanish. For $\mu =t$ this immediately givesFor a spatial index $\mu =i$,by the stated property. Thus the perturbation is transverse.

Its Minkowski trace is purely spatial:Adding the displayed components and collecting the coefficients of the independent functions $A$, $B$, and $C$ makes each coefficient vanish separately, soTogether with $h_{t\mu }=0$ and ${\partial }^{\nu }{h}_{\mu \nu }=0$, this proves that the wave is in transverse-traceless gauge.