Is there a way to prove that the Kronecker delta $\delta_{ij}$ is indeed the only isotropic second order tensor (i.e. invariant under rotation), i.e. so we can write $T_{ij} = \lambda \delta_{ij}$ for some constant $\lambda$?
By rotational invariance I mean: $$ T_{ij} = T^\prime_{ij} = R_{ip} R_{jq} T_{pq}\text{,} $$ where the matrices $R_{ij}$ are orthogonal.
It is very straightforward to show that $\delta_{ij}$ is invariant, but how can I show that it is unique?
Step 1: Realize $T$ is diagonal. Let $$R_{kl}=\begin{cases} -1 & \text{if }k=l=j\\ \delta_{kl} & \text{otherwise}\\ \end{cases}$$ therefore $R$ is the reflection w.r.t to the hyperplane perpendicular to the j-th vector in the standard ordered basis. $$R=R^T\land R^2=I\Rightarrow R^TR=RR^T=I$$ Therefore: $$T_{ij}=\sum_{p,q}R_{ip}R_{jq}T_{pq}=R_{ii}R_{jj}T_{ij}\\ i\neq j\Rightarrow T_{ij}=-T_{ij}\Rightarrow T_{ij}=0$$
Step 2: Realize $T_{jj} = T_{11}$. Let $$P_{kl}=\begin{cases} \delta_{jl} & \text{if } k=1\\ \delta_{1l} & \text{if } k=j\\ \delta_{kl} & \text{otherwise} \end{cases}$$ therefore $P$ is the permutation matrix that interchanges the 1st and j-th rows on left multiplication. $$(P^TP)_{kl}=\sum_{m}P^T_{km}P_{ml}=\sum_{m}P_{mk}P_{ml}=\sum_{m\neq1,j}P_{mk}P_{ml}+\sum_{m=1,j}P_{mk}P_{ml}\\ =\sum_{m\neq1,j}\delta_{mk}\delta_{ml}+\delta_{jk}\delta_{jl}+\delta_{1k}\delta_{1l}=\sum_{m}\delta_{mk}\delta_{ml}=\delta_{kl}$$ Therefore: $$T_{jj}=\sum_{p,q}P_{jp}P_{jq}T_{pq}=\sum_{q}P_{jq}^2T_{qq}=\sum_{q}\delta_{1q}^2T_{qq}=\sum_{q}\delta_{1q}T_{qq}=T_{11}$$