Then show that $v_1,v_2,\ldots v_n$ are linearly independent for $n=2011$. Let $n\ge 3$ be an integer and let $u_1,u_2,\ldots ,u_n$ be $n$ linearly independent elements in a vector space over $\Bbb R$.Set $u_0=0,u_{n+1}=u_1$.
Define $v_i=u_i+u_{i+1}$ and $w_i=u_{i-1}+u_i$ for $i=1,2,\ldots n$
Then show that $v_1,v_2,\ldots v_n$ and $w_1,w_2,\ldots w_n$ are linearly independent for $n=2011$.
I am unable to understand how to show this for $n=2011$ .Please help.
 A: Instead of following the definition(a good way), my idea is:
The transformation mapping $\{u_i\}$ to $\{v_i\}$ is:
\begin{bmatrix}
1 & 1 & 0 &0&...& 0\\
0 & 1 & 1 &0&...&0\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\
1&0&...&...&0&1
\end{bmatrix}
and the transformation mapping $\{u_i\}$ to $\{w_i\}$ is:
\begin{bmatrix}
1 & 0 & 0 &0&...& 0\\
1 & 1 & 0 &0&...&0\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\
0&0&...&...&1&1
\end{bmatrix}
and we can check if $\{v_i\}$ and $\{w_i\}$ are independent by checking the transformations invertible
(for $\{v_i\}$ the condition $n=2011$ should be used).
A: You do not need $n=2011$, you just need $n$ to be odd.
Let $\lambda_1,\ldots,\lambda_n$ be scalars such that $\sum_{k=1}^n \lambda_k v_k=0$,
we must show that all the $\lambda_k$ are zero. Now, if we put $\lambda_{0}=\lambda_n$,
$$
\sum_{k=1}^n \lambda_k v_k=\sum_{k=1}^n \lambda_k u_k+\sum_{k=1}^n \lambda_k u_{k+1}
=\sum_{k=1}^n \lambda_k u_k+\sum_{k=1}^{n} \lambda_{k-1} u_{k}=
\sum_{k=1}^n (\lambda_k+\lambda_{k-1}) u_k 
$$
Since the $(u_k)$ are linearly independent, we have $\lambda_k+\lambda_{k-1}=0$ for
$1\leq k \leq n$. So $\lambda_{k+r}=(-1)^r \lambda_k$ by induction on $r$, so 
$\lambda_{n}=(-1)^{n-1} \lambda_1$, as $n$ is odd we deduce that $\lambda_n=\lambda_1$ and $0=\lambda_1+\lambda_n$ then gives $\lambda_1=\lambda_n$, so that all the $\lambda_k$ must be zero.
