= Solution
Let $R_\theta$ be rotation about the third axis, so invariance means $R_\theta T R_\theta^T=T$. At $\theta=\pi$, its diagonal signs are $(-1,-1,1)$; each mixed entry $T_{13},T_{23},T_{31},T_{32}$ changes sign and must vanish.
For the remaining planar block $M=\begin{pmatrix}p&q\\r&s\end{pmatrix}$, the quarter-turn $J=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ gives
$$
JMJ^T=\begin{pmatrix}s&-r\\-q&p\end{pmatrix}=M.
$$
Hence $s=p$ and $r=-q$. Writing the constants as $\alpha,\omega,\beta$ gives
$$
\boxed{T=\begin{pmatrix}\alpha&\omega&0\\-\omega&\alpha&0\\0&0&\beta\end{pmatrix}.}
$$
Conversely, $M=\alpha I-\omega J$ commutes with every planar rotation, so every displayed matrix is invariant under every rotation about the axis. This is an <oriented-axis invariant Cartesian tensor>. Symmetry was not assumed, so the antisymmetric parameter $\omega$ need not vanish.
Back to article page
