Find the smallest number of Linearly Dependent Vectors. I am stuck in the following problem:
Consider the vector space $\Bbb R^{2016}$ over the field $\Bbb R$. Then,
"What is the smallest
positive integer $k$ for which the following statement is true: given any $k$ vectors
$v_1, . . . , v_k ∈ $$\Bbb R^{2016}$, there exist real numbers $a_1, . . . , a_k$, not all zero, such that
$a_1v_1 + · · · + a_kv_k$ = $0$ and $a_1 + · · · + a_k$ = $0$."
any help would be appreciated...
 A: Firstly, notice that $k>2016$. If $k\leq2016$ then you may pick $v_1,v_2,\cdots,v_k$ linearly independent. Hence, you cannot possibly have 
$a_1v_1+a_2v_2+\cdots+a_k v_k=0$ where not all $a_i$ are zero.
Now, if $k=2017$ take $e_i$ $1\leq i \leq 2017$ the canonical base, and $u=\sum_{i=1}^{2016}e_i$. Therefore, 
$$\sum_{i=1}^{2016}b_ie_i +bu = \sum_{i=1}^{2016}(b_i+b)e_i=0$$
implies that $b_i=-b,\forall i$. So $\sum_{i=1}^{2016}b_i +b= -2015b$. So, the require condition fails.
For $k=2018$, pick any $w_1,w_2,\cdots,w_{2018}$ define $f:\mathbb{R}^{2018} \rightarrow \mathbb{R}^{2016}$ as follows. $f(e_i) =w_i $. The kernel of $f$ has dimension at least $2$. Let $v_1,v_2 \in \text{Ker}( f)$ be two linearly independent vectors. 
Now, define $g:\mathbb{R}^{2018} \rightarrow \mathbb{R}$ $g(e_i)=1$. And take $h=g_{|<v_1,v_2>}$ the restriction of $g$ on the space spanned by $v_1,v_2$. Since $<v_1,v_2>$ has dimension $2$ the kernel of $h$ is non trivial let's say it is spanned by $w=l_1v_1+l_2v_2$. Then $g(w)$ is the sum of the $a_i$ which is equal to zero, and also $f(w)$ is a non trivial sum of the $w_i$. 
So we conclude.
