= Solution
For a <Cartesian tensor> under a proper orthogonal change $R$, the rank-$n$ law is $T'_{i_1\cdots i_n}=R_{i_1j_1}\cdots R_{i_nj_n}T_{j_1\cdots j_n}$, with repeated indices summed. An <isotropic tensor> has $T'=T$ for every such rotation.
For rank two, $T=RTR^T$. A half-turn about a coordinate axis, obtained by applying a quarter-turn twice, reverses each off-diagonal component with exactly one index on that axis. Invariance forces every off-diagonal entry to vanish. A quarter-turn about the third axis exchanges $T_{11}$ and $T_{22}$; quarter-turns about the other axes equate them with $T_{33}$. Thus the <isotropic second-rank tensor> is
$$
\boxed{T_{ij}=\lambda\delta_{ij}.}
$$
Conversely $R(\lambda I)R^T=\lambda I$, proving sufficiency.
Back to article page
