Why does circular symmetry of complex envelopes mean that $\mathbb{E} \Big[\mathbf{x}[i] \mathbf{x}^T[i] \Big] = \mathbf{0}$? I am making my way through this paper here, and I understand everything up to equation 16, and I am stuck here. 

Long story short, the vector $\mathbf{x}[i]$ here is an $N$ by 1 vector, consisting of the complex envelopes of signal + noise + interference across $N$ different sensors, taken at some time index $i$. We have a desired signal, an interference signal, and AWGN noise at each sensor. Naturally, the desired and interference signals at each sensor are delayed versions of each other, but the AWGN at each sensor is independent of the AWGN at another sensor.

What I do not understand, is the author's claim, where he states that since the desired and/or interference signals are just delayed copies of each other, "the circular symmetry of the complex envelopes thus implies that:"

$$
\mathbb{E} \Big[\mathbf{x}[i] \mathbf{x}^T[i]  \Big] = \mathbf{0}
$$

Circular symmetry of the complex envelopes? What does that mean, and why do we get the above equation from that?
A complex random variable with dimension being 2 $Z=X+jY$ can be defined as a Random vector with components $X$ and $Y$ which take values along 2 different dimensions. The Probability Density Function of $Z$ is defined using the PDFs of $X$ and $Y$. The Expectation is defined as $\mathrm{E}(Z)=\mathrm{E}(X)+j\mathrm{E}(Y)$ and the variance is defined the usual way as $Z$ being a complex function of $X$ and $Y$ and the Variance is $\mathrm{VAR}(Z)=\mathrm{E}(|Z^2|)-|\mathrm{E}(Z)|^2$. Intuitively speaking, $Z$ is a complex function of two random variables which just assumes value $X+jY, \forall x,\forall y$. The PDF is written as follows: Write the 2-dimensional Random variable as $$\tilde{Z}=\left[\begin{array}{ccc}X \\Y\end{array}\right]$$where $\tilde{Z}$ is a Real random vector. The PDF of $Z$ is the PDF of $\tilde{Z}$.
Now with Complex Random vectors, $\mathbf{Z}=\mathbf{X}+j\mathbf{Y}$ is a vector of Complex Random variables which has a lot of interesting properties in a sense that it has more to offer than a Vector of Real Random Variables. Particularly of importance are the Covariance, Pseudocovariance(this is something new) and the Expectation. In our situation, we have a special Complex Random vector called the Symmetric  Random Vector $\mathbf{Z}$.
Define the $2n$-dimensional Complex Random Vector as $$\tilde{\mathbf{Z}}=\left[\begin{array}{ccc}\mathbf{X} \\\mathbf{Y}\end{array}\right]$$ where $\mathbf{X}$ and $\mathbf{Y}$ are $n$-dimensional real random vectors. The PDF definition follows the same from Complex Random Variable. You can look up what the Pseudocovariance and Variance mean for a Complex Random Vector. I'm referring to Digital Communications by Proakis here in my definitions and notations. 

Now as in the comments, if $\mathbf{Z}$ is a Symmetric i.e the PDF doesn't change on rotation, we can say that $\mathrm{E}(\mathbf{ZZ^\mathrm{T}})=0$. This is a consequence of the following. Since $\mathbf{Z}$ and $\mathbf{Z}e^{j\theta}$ have the same PDF, $$\mathrm{E}(\mathbf{Z})=\mathrm{E}(e^{j\theta}\mathbf{Z})=e^{j\theta}\mathrm{E}(\mathbf{Z}), \forall \theta$$ If we had $\theta=\pi$ (the condition has to be satisfied for all $\pi$), we can see that $$\mathrm{E}(\mathbf{Z})=\mathrm{E}(-\mathbf{Z})=-\mathrm{E}(\mathbf{Z})$$ which clearly says that $\mathrm{E}(\mathbf{Z})=0$. Going further, we can prove that $\mathrm{E}(\mathbf{Z}\mathbf{Z}^\mathrm{T})=0$ the same way assuming a special case of $\theta=\frac{\pi}{2}$. And hence Circularly Symmetric Complex Random Vectors are Zero-mean and Proper. The last equation left unproven here is the definition of being proper.