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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.
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.
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).
