Variance of a dynamical system

Question

The Problem

Considering the given system $$\dot{v_i}(t) = \sum_{j=1}^N \sigma_{ij} (v_j(t) - v_i(t)) + u_i(t), \quad i=1...N$$ where $v_i(t), u_i(t) \in \mathbb{R}^d$ and $\sigma_{ij} = \sigma_{ji} \geq 0$.

I want to show two things:

1. Let $\bar{v}(t) = \frac{1}{N} \sum_{i=1}^N v_i(t)$ and $\bar{u}(t) = \frac{1}{N} \sum_{i=1}^N u_i(t)$. Show that $$\dot{\bar{v}}(t) = \bar{u}(t)$$
2. Define the variance by $V(t) = \frac{1}{2N^2} \sum_{i,j=1}^N \|v_i(t) - v_j(t)\|^2$. Show that $$V(t) = \frac{1}{N} \sum_{i=1}^N \|v_i(t) - \bar{v}(t)\|^2$$ and $$\dot{V}(t) = - \frac{1}{N} \sum_{i,j=1}^N \sigma_{ij} \|v_i(t) - v_j(t)\|^2 +\frac{2}{N}\sum_{i=1}^N \langle v_i(t) - \bar{v}(t), u_i(t) \rangle$$ with $\| \cdot\|$ as the euclidean norm in $\mathbb{R}^d$ and $\langle \cdot,\cdot \rangle$ the associated inner product.

What I tried so far

1. I tried to write the sum out and it became pretty clear, that one has always a term $\sigma_{ij}(v_i(t) - v_j(t))$ and later on in the sum one has $\sigma_{ji}(v_j(t) - v_i(t))$ which cancel out. And the terms for $i=j$ obviously vanish. But I would really like to solve this, without writing the sum explicitly.
2. For the first problem I tried to use some sort of binomial formula like $\|a - b\|^2 = \|a\|^2 - 2 \langle a, b\rangle + \|b\|^2$ and using the explicit form of $\bar{v}(t)$, but there two main problems occured to me
   • There is somehow a factor 2 which I cannot get rid of.
   • The last term after I applied the binomial formula looks like a sum of norms, while in the expression I want to get, I have a norm of a sum, which are obviously not the same.