Let $v_1,\ldots, v_n$ be linear independent vectors and $v$ a linear combination of those, i.e. there are $\beta_1,\ldots,\beta_n\in\mathbb{R}$, such that $v=\sum_{j=1}^n\beta_jv_j$. However, $v\ne v_j$ for any $j\in\{1,\ldots, n\}$.
Is it true, that $v_1-v,\ldots, v_n-v$ are still linear independent?
For $\alpha_1,\ldots, \alpha_n\in\mathbb{R}$, I find
$$0=\sum_{j=1}^n\alpha_j(v_j-v)=\sum_{j=1}^n\alpha_jv_j-(\sum_{j=1}^n\alpha_j)v$$ such that
$$(\sum_{j=1}^n\alpha_j)\sum_{k=1}^n\beta_kv_k=\sum_{k=1}^n\alpha_kv_k$$ and
$$\alpha_k=(\sum_{j=1}^n\alpha_j)\beta_k$$
Can I deduce from this, that $\alpha_k=0$?