 Solution

Use the quarter-turn rotation matrixFor a general second-order tensor, direct multiplication givesRequiring $RT{R}^T=T$ gives $T_{11}=T_{22}$, $T_{12}=-T_{21}$, $T_{13}=-T_{23}$ together with $T_{23}=T_{13}$, and $T_{31}=-T_{32}$ together with $T_{32}=T_{31}$. Thus all mixed axial-transverse entries vanish andFor an arbitrary rotation angle $\theta$, the transverse rotation is $Q_{\theta }=\cos \theta I+\sin \theta J$, where $J=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$. The transverse block of $T$ is $\alpha I-\omega J$, so it commutes with $Q_{\theta }$. Therefore $Q_{\theta }(\alpha I-\omega J)Q_{\theta }^T=\alpha I-\omega J$, while the axial component $\beta$ is unchanged.

Hence invariance holds under every rotation about the axis, not just the quarter-turn. This is the general axisymmetric second-order tensor; a symmetric tensor would additionally require $\omega =0$, but that restriction was not assumed.