Linearly independent subset - a simple solution? 
Problem: Let $\{ v_1, \ldots, v_n \}$ be a linearly independent subset of $V$, 
  a vector space. let  $$ v= t_1 v_1 + \cdots + t_n v_n $$  where $t_1,
 \ldots t_n \in \mathbb{R}$. For which $v$ is the set $\{v_1 + v ,\ldots , v_n + v \}$ linearly independent?

My ideas were: 
Firstly, any linear combination of the new set is, 
$$ \sum \lambda_i (v_i + v) = \sum \lambda_i v_i + sv = \sum (\lambda_i + st_i) v_i = 0 $$ 
where $s = \sum \lambda_i$. By linear independence of $v_i$, we must have 
$\lambda _i + st_i = 0$ for $i=1, \ldots , n$. 
Summing up yields $ s + s \sum t_i = 0 \Rightarrow s( 1 + \sum t_i) = 0 $. 

My claim is: $\sum t_i \not = -1 \Leftrightarrow$ set is linearly
  independent. 
Proof:  If $\sum t_i \not = -1$, then $s = 0$, hence $\lambda _i = 0$ for $i=1, \ldots, n$. So the new set is linearly independent. 
On the other hand if $\sum t_i = -1$, then regarding the original
  system of linear equations, where $$\lambda_i + st_i = \sum_{i \not=j}
 \lambda_j t_i + \lambda_i (1+t_i) = 0$$ for $i-1, \ldots, n$. We wish
  to find a non trivial set of solutions for $i=1, \ldots, n$. Written
  out in matrix form, we have a matrix with all column sums equal to $0$
  (column $i$ has sum $\sum t_i +1 = 0$), hence the transpose of this
  matrix, has a non trivial solution, eigenvector $(1, \ldots, 1 )^T$.
Therefore, the transpose matrix is not invertible - hence, the matrix
  itself is not invertible. So there exists a non trivial solution.

I think this solution is a bit too long winded - I am hoping if anyone could give a more elmentary solution that does not use matrices.  Thank you! 

It is easy to explicitly guess a nontrivial solution for the system of equations $$\lambda_i + st_i = 0$$ By setting $\lambda_i = t_i$ for all $i$, we get $\lambda_i + st_i = t_i-t_i = 0$ for all $i$ since $s=-1$ by assumption. Also since $\sum t_i = -1$, atleast one $t_i$ is nonzero, so the solution is nontrivial.
