 Solution

For a Cartesian change of components $x_i^{\prime }=R_{ij}{x}_j$, a rank-$n$ tensor transforms according towith repeated indices summed. An isotropic tensor has unchanged components under every proper rotation matrix.

For rank two write $T^{\prime }=RT{R}^T$. Invariance under quarter-turns implies invariance under their squares, the coordinate half-turnsEach off-diagonal entry changes sign under at least one of these, so all off-diagonal entries vanish. A quarter-turn about the third axis then interchanges the first two diagonal entries, making them equal; a quarter-turn about the first axis equates the other two. ThereforeConversely $R(\lambda I)R^T=\lambda I$ for every rotation. This proves the isotropic second-rank tensor classification without assuming that the original tensor was symmetric. Also $\lambda =T_{kk}/3$ is a scalar, since the trace is rotation-invariant.