# Orthogonality of Gravitational Wave Polarizations

Here is a question from Schutz's GR book: My (short-lived) attempt at a solution:

I define the direction of propagation of the waves as the z direction. $$(A^{\mu\nu})^*B_{\mu\nu}=2(A^{xx})^*B_{xx}+2(A^{xy})^*B_{xy}=0 \text{ (TT guage)}$$ $$(B_{\mu\nu})=B_{xx}\begin{pmatrix} 1 & -(A^{xx})^*/(A^{xy})^*\\ -(A^{xx})^*/(A^{xy})^* & -1\\ \end{pmatrix} = \frac{B_{xx}}{(A^{xy})^*}\begin{pmatrix} (A^{xy})^* & -(A^{xx})^*\\ -(A^{xx})^* & -(A^{xy})^*\\ \end{pmatrix}$$

At this point, I'm not really sure what to do. I don't know what it means to "rotate" a matrix so I assume the correct thing to do is complex rotate each component, but I'm not entirely sure what this entails either. I tried multiplying each component by $e^{i\pi/4}=(1+i)/\sqrt{2}$ but I'm not sure how this helps. Any advice?

Rank 2 tensors transform under rotations like $$B_{\mu'\nu'}=\Lambda^\mu_{\mu'}\Lambda^\nu_{\nu'}B_{\mu\nu}$$ where $\Lambda^\mu_{\mu'}$ is the rotation matrix $\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \\ \end{pmatrix}$. Using $\theta=\pi/4$ will yield $B^{\mu'\nu'}\propto (A^{\mu\nu})^*$.