Is {$u_{1},u_{2},\cdots,u_{k},v$} linearly independent? Question:
Given {$u_{1},u_{2},\cdots,u_{k},u_{k+1}$} is a set of linear independent vectors in $R^n$ 
and $v = u_{1}+u_{2}+\cdots+u_{k}+u_{k+1}$, 
prove that {$u_{1},u_{2},\cdots,u_{k},v$} linearly independent. 
My solution:
My definition of {$u_{1},u_{2},\cdots,u_{k},v$} is 
$$ \lambda_1u_1+\dots+\lambda_ku_k+\lambda_{k+1}v=0 $$, where  $\lambda_i$ (with $i\in\{1,\dots,k+1\}$. 
Suppose that $\lambda_{k+1}=0$, this would mean that $\{u_1,\dots,u_k\}$ must be linearly dependent, which is contradictory. 
I probably messed up my concepts. Please advice, thanks. 
 A: Hint:
$$ \lambda_1u_1+\dots+\lambda_ku_k+\lambda_{k+1}v=0 $$
$$\implies \lambda_1u_1+\dots+\lambda_ku_k+\lambda_{k+1}\left(u_{1}+u_{2}+\cdots+u_{k}+u_{k+1}\right)=0$$
$$\implies (\lambda_1+\lambda_{k+1})u_1+\dots+(\lambda_k+\lambda_{k+1})u_k+\lambda_{k+1}u_{k+1}=0$$
What can you conclude from the last equation?
A: From the given statement, we know that the only scalars, $\alpha_i \in \mathbb{R}$, such that $$\alpha_1 u_1 + \alpha_2 u_2 + \cdots + \alpha_{k+1} u_{k+1} = 0 \tag{*}$$ are $\alpha_1 = \alpha_2 = \cdots = \alpha_{k+1}=0 $. 
Now consider $$\beta_1 u_1 + \beta_2 u_2 + \cdots + \beta_k u_k + \beta_{k+1} v = 0$$ $$\Rightarrow \beta_1 u_1 + \beta_2 u_2 + \cdots + \beta_k u_k + \beta_{k+1} (u_1 + u_2 + \cdots + u_{k+1}) = 0$$ $$\Rightarrow (\beta_1 + \beta_{k+1})u_1 + \cdots + (\beta_k + \beta_{k+1})u_k + \beta_{k+1} u_{k+1} = 0$$ Notice this equation is the same as $(*)$, with $\alpha_1 = \beta_1 + \beta_{k+1}$,$\cdots$,  $\alpha_k = \beta_k + \beta_{k+1}$, $\alpha_{k+1} = \beta_{k+1}$. 
We know that all the $\alpha_i=0$, so $\beta_1 + \beta_{k+1} = 0, \beta_2 + \beta_{k+1}=0, \cdots, \beta_k + \beta_{k+1}=0, \beta_{k+1}=0$ and $\{u_1, u_2, \cdots, u_k, v\}$ is thus linearly independent. 
A: Note that a linear combination of the second set's vectors is always a linear combination of the first set's vectors (via the distributive property).
Linear independence is a condition on linear combinations of a set. So if the condition holds for linear combinations of the first set - and it does, by hypothesis - it also holds for linear combinations of the second set.
