Here is an elementary solution:
Let $a_1=(\cos(\theta), \sin(\theta))$ then $a_i= \cos(\theta+\frac{2(i-1)\pi}{n}), \sin(\theta+\frac{2(i-1)\pi}{n})$ thus
$$a_i a_i^T= \left( \begin{array}{c c} \cos^2(\theta+\frac{2(i-1)\pi}{n}) & \cos(\theta+\frac{2(i-1)\pi}{n})\sin(\theta+\frac{2(i-1)\pi}{n}) \\ \cos(\theta+\frac{2(i-1)\pi}{n})\sin(\theta+\frac{2(i-1)\pi}{n}) & \sin^2(\theta+\frac{2(i-1)\pi}{n}) \\ \end{array} \right)$$
Now the result follows imediatelly from the simple trig identities
$$\sum_{i=1}^n \cos^2(\theta+\frac{2(i-1)\pi}{n})= \sum_{i=1}^n \sin^2(\theta+\frac{2(i-1)\pi}{n}) =n/2$$
$$\sum_{i=1}^n \cos(\theta+\frac{2(i-1)\pi}{n})\sin(\theta+\frac{2(i-1)\pi}{n})=\frac{12}{2} \sum_{i=1}^n \sin(2\theta+\frac{4(i-1)\pi}{n})=0 \,.$$
The last question in the problem is equivalent to the following:
Does
$$\sum_{i=1}^n \cos^2(\theta_i)= \sum_{i=1}^n \sin^2(\theta_i) =n/2$$
$$ \sum_{i=1}^n \sin(2\theta_i)=0 \,.$$
imply
$$ \sum_{i=1}^n \sin(\theta_i)=\sum_{i=1}^n \cos(\theta_i)=0 \,?$$
P.S. Can anyone fix my matrix?
