Proving linear indepence under certain conditions. While revising for my linear algebra paper in a few days, I came across this question in a past paper:
Let n, k $\in \mathbb N, v_1...v_n \in \mathbb R^k$ and $w_i = \sum_{j=1}^i v_j$ for  $i=1... n.$ Show that $v_1...v_n$ is linearly independent, if and only if $w_1,...w_n$ are linearly independent.
If I'm not wrong I'd need to show that the assumption holds in both directions, i.e 
1) $w_1,...w_n$ are linearly independent when $v_1...v_n$ are linearly independent.
2) $v_1...v_n$ are linearly independent when $w_1,...w_n$ are linearly independent.
I've not had many ideas on how to go about this question, but I thought of using the definition of linear independence so none of the vectors can be defined as  as a linear combination of the others. I haven't, however, been able to make proper use of this definiton in my attempts at proving the above.
I hope I can get some ideas on how to go about proving the above statement. Thanks in advance!
 A: For the case $k=n$ the simplest thing would be to write the determinant that has each vector as column and prove that the one made up by the $w_i$ is nonzero $\Leftrightarrow$ the one with the $v_j$ is too. The determinant being nonzero implies the columns are independent.
For the case $n<k$ I'd work on each implication separately and argue with reductio ad absurdum, that is, supposing they are not independent and come to a contradiction.
Hope it helps!
A: Assume $n\leq k$.
Assume $w_1,w_2,\dots,w_n$ are linearly dependent, i.e. Let $\sum_{i=1}^{n} a_i\cdot w_i = 0$ for some $a_i \neq 0$. Then 
\begin{eqnarray}
0 & = & a_1v_1 + a_2(v_1+v_2)+a_3(v_1+v_2+v_3) + \cdots  \\
& = &(a_1+a_2+\cdots +a_n)v_1 + (a_2+\cdots + a_n)v_2+ \cdots + a_nv_n \\
& = & b_1v_1+b_2v_2+\cdots + b_nv_n.
\end{eqnarray}
Since some $a_i\neq 0$ then there must be some $b_j\neq 0$ and therefore $v_1,v_2,\cdots,v_n$ are linearly dependent. 
Assume $v_1,v_2,\cdots,v_n$ are linearly dependent, i.e. $\sum_{i=1}^nb_iv_i = 0$ for some $b_i \neq 0$. Since $w_i = \sum_{j=1}^iv_i$ then $v_j = w_j-w_{j-1}$ for $j\geq 2$ and $v_1 = w_1$. Then
\begin{eqnarray}
0 & = & b_1v_1 + b_2v_2 + \cdots + b_nv_n\\
& = &  b_1w_1 + b_2(w_2-w_1) + \cdots + b_n(w_n-w_{n-1})\\
& = & (b_1-b_2)w_1 + (b_2-b_3)w_2 + \cdots + b_nw_n \\
& = & a_1w_1 + a_2w_2 + \cdots + a_nw_n
\end{eqnarray}
Since some $b_i \neq 0$ then there must be some $a_j\neq 0$ and therefore $w_i$, $1\leq i\leq n$, must be linearly dependent.
