Why is the dipolar spin-spin interaction matrix diagonalizable and traceless? In this Wikipedia article on zero field splitting, the Hamiltonian is defined as
$$H=SDS$$
where $S$ is the total spin $S_{1}+S_{2}$ and $D$ is the dipolar spin-spin interaction matrix. It is asserted that $D$ is diagonalizable and traceless. But why should it be so?
 A: It is straight forward to show that $D$ is diagonalizable:

*

*If the first spin points in $x$-direction and the second spin points in $y$-direction the interaction energy of the two dipoles is zero.

*There is nothing special about the perpendicular direction pairs $\{\vec e_x, \vec e_y\}$. Therefore, the same argument applies to all perpendicular direction pairs $\{\vec e_i, \vec e_j\}$ with $i\ne j$. Thus we conclude that the matrix element $D_{ij}=0$, if $i\ne j$. This immediately implies that the dipole-dipole interaction matrix $D$ is diagonalisable.

I don't know a simple mathematical argument to show that the trace of the dipole-dipole interaction Hamiltonian
$$
H \propto \frac{\bf S_1 \cdot S_2}{r^3} - 3 \frac{({\bf S_1 \cdot  r}) ({\bf S_2 \cdot r})}{r^5}
$$
is zero. Personally, I am happy to accept this by noting that the dipole-dipole interaction becomes zero if it is averaged over all possible orientations. However, people have done this calculation, see e.g. this book on page 49, which I also append as a screen shot:

