Showing $E[\|x\|^2xx']=\text{tr}(\Sigma)\Sigma+2\Sigma^2$

Question

Empirically the following seems to hold for $x \sim \text{Normal}(0, \Sigma)$, any idea how to show this rigorously?

$$E[\|x\|^2xx']=\text{tr}(\Sigma)\Sigma+2\Sigma^2$$

Answer 1

OK, with $\Sigma$ diagonal the result is also diagonal and $i,i$th entry of that matrix is $$E[\|x\|^2 x_i^2]=\sum_j E[x_i^2 x_j^2]$$ The $i$th term of that sum is Gaussian 4-th moment, so $3\sigma_i^2$ while remaining terms factor as $\sigma_i \sigma_j$ because of independence. Rearranging we get

$$E[\|x\|^2 x_i^2]=\text{tr}(\Sigma) \sigma_i + 2\sigma_i^2 $$