In my reader it is stated that $\langle \mathbf r\mathbf r^T\rangle=\tfrac 1 3\langle r^2\rangle $ for isotropic vectors $\mathbf r$ but with no further explanation and I'd like to know where it comes from. Here $\langle \ \rangle $ denotes an expectation value, $\mathbf r\in\mathbb R^3$ is a position vector and is the $3\times 3$ identity matrix.
My first guess would be to split the tensor in an spherically symmetric part and a non-symmetric traceless part: \begin{align} \mathbf r\mathbf r^T &=\tfrac 1 3\text{tr}(\mathbf r\mathbf r^T )+\big(\mathbf r\mathbf r^T-\tfrac 1 3\text{tr}(\mathbf r\mathbf r^T )\big) \\ &=\underbrace{\tfrac 1 3(\mathbf r\cdot\mathbf r)}_{\text{spher. symmetric}}+\underbrace{\big(\mathbf r\mathbf r^T-\tfrac 1 3(\mathbf r\cdot\mathbf r)\big)}_{\text{non-sym. traceless}} \end{align} My second guess would be that taking the expectation value removes the non-symmetric part but I have trouble showing this. Also I don't understand why the non-symmetric part has to be traceless so in turn I dont understand see where the prefactor $\tfrac 1 3\langle r^2 \rangle$ comes from. I could have also picked a more general split $a+(\mathbf r\mathbf r^T-a)$. Any help?
An answer to the first question: let $A = \langle \mathbf r\mathbf r^T \rangle$; note that $A$ is necessarily symmetric. We note that for any unit vector $\mathbf v$, we have $$ \langle (\mathbf r ^T \mathbf v)^2 \rangle = \langle \mathbf v^T \mathbf r \mathbf r^T \mathbf v\rangle = \mathbf v^T \langle \mathbf r \mathbf r^T \rangle \mathbf v = \mathbf v^T A\mathbf v. $$ Now, because $\mathbf r$ is isotropically distributed, $\mathbf v^TA\mathbf v$ must be equal to some constant $c$ for all unit-vectors $\mathbf v$. Now, for a symmetric matrix $M$: if $\mathbf v^T M \mathbf v = 0$ holds for all unit vectors $\mathbf v$, then it must hold that $M = 0 $. Thus, we may conclude that $A - c I = 0$ (where $I$ denotes the identity matrix). Thus, $A = cI$.
Now, note that $$ \operatorname{Tr}\langle \mathbf r \mathbf r^T \rangle = \langle r^2\rangle = \operatorname{Tr}(c I) = 3c \implies c = \frac 13 \langle r^2 \rangle. $$