Linearly dependent system of vectors with sum of coefficients $0$ Let $V$ be a vector space over a field $K$ so that $dim(V) = n$. Let $e_1, e_2,\dots,e_{n+2}$ be a vector system. Prove that there exists $k_1,k_2,\dots,k_{n+2} \in K$ which are not all $0$ so that
$$
\sum_{i=1}^{n+2} k_ie_i = 0 \quad \sum_{i=1}^{n+2}k_i=0
$$
My attempt was that since there are more than $n$ vectors in the system and every basis of this vector space must have exactly $n$ vectors, then this system is linearly dependent which means that there exists such $k_i$ which are not all $0$ that 
$$
\sum_{i=1}^{n+2} k_ie_i = 0
$$
I have trouble proving the second part though. How can I be show that the sum of coefficients can also be $0$ for some $k_i$?
 A: Since $n+1>\dim(V)$, there exists $a_j$ for $j=1,2,\ldots,n+1$ not all zero (WLOG suppose $a_{n+1}\neq 0$) such that $$\sum_{j=1}^{n+1} a_je_j=0$$ and $b_j$ for $j=1,2,\ldots,n,n+2$ not all zero such that $$\sum_{j=1\\j\neq n+1}^{n+2} b_je_j=0.$$ Set $a_{n+2}:=0$ and $b_{n+1}:=0$ so that $$\sum_{j=1}^{n+2} a_je_j=0=\sum_{j=1}^{n+2} b_je_j.$$ Now set $$a:=\sum_{j=1}^{n+2} a_j \qquad b:=\sum_{j=1}^{n+2} b_j.$$ Now take $k_j:=ba_j-ab_j$ for $j=1,2,\ldots,n+2$. Notice that $k_{n+1}=ba_{n+1}\neq 0$ (unless $b=0$, but if this is the case then we're done).
A: Let $M$ be the $n\times (n+2)$ matrix whose first column is $e_1$, second column is $e_2$, ... Then let $A$ be the $(n+1)\times (n+2)$ matrix obtained by adding to the bottom of $M$ an extra row $[1, 1, ... 1]$. Since $A$ has more columns than rows, there is a vector $k=[k_1, ...,k_{n+2}]^T\neq 0$ so that $Ak=0$ (any linear map $A$ from ${\mathbb R}^{n+2}$ to ${\mathbb R}^{n+1}$ cannot be one-to-one, so must have nontrivial kernel - i.e. some $k\neq 0$ with $Ak=0$). Now $k_1, k_2, ...,k_{n+2}$ is just the numbers needed here.
