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Let $n\geq3$ be an integer, let $u_1,u_2,u_3,\ldots,u_n$ be $n$ linearly independent elements over a vector space over $\mathbb{R}$. Set $u_0=0$ and $u_{n+1}=u_1$ and define $v_i=u_i+u_{i+1}$ and $w_i=u_{i-1}+u_{i}$ for $i=1,2,\ldots,n$, then

  1. $v_1,v_2,v_3,\ldots,v_n$ are linearly independent if $n=2010$.

  2. $v_1,v_2,v_3,\ldots,v_n$ are linearly independent if $n=2011$.

  3. $w_1,w_2,w_3,\ldots,w_n$ are linearly independent if $n=2010$.

  4. $w_1,w_2,w_3,\ldots,w_n$ are linearly independent if $n=2011$.

I am stuck on this problem. Can anyone help me please?

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Hint that should illustrate the general line of thought: $$v_1-v_2+v_3-v_4+ \cdots \pm v_k \\= u_1 +u_2 - (u_2 + u_3) + (u_3 + u_4) - \cdots \pm (u_{k}+ u_{k+1}) = u_1 \pm u_{k+1},$$ the sign depending on whether $k$ is even or odd.

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  • $\begingroup$ @Prasanta: Let us start with the case 1) and $k = n = 2010$. Then the last sign will be a $-$, and the alternating sum will be equal to $u_1 - u_{n+1} = u_1 - u_1 = 0$. What does that say about the independency of the $v_i$? Case 2) is similar, and here you can show that the $v_i$ are independent by using the definition. If you write $$\sum_{i=1}^n a_i v_i = \sum_{i=1}^n a_i(u_i+u_{i+1}) = 0,$$ then you can deduce from the independency of the $u_i$, that $a_1 + a_2 = 0$, $a_2 + a_3 = 0$, up to $a_{n-1}+a_n = 0$, and $a_1 + a_n = 0$. Is there a non-zero solution when $n$ is odd? If $n$ is even? $\endgroup$ – fuglede Jun 22 '13 at 13:49
  • $\begingroup$ @ fuglede,case 1) n=2010. $v_i$ are inearly dependent... $\endgroup$ – user45799 Jun 22 '13 at 14:01
  • $\begingroup$ @Prasanta: Yep! $\endgroup$ – fuglede Jun 22 '13 at 14:03
  • $\begingroup$ @ fuglede, when n is odd $a_1+a_n=0 $, Since $u_i$ are L.I., case 2) should be right... $\endgroup$ – user45799 Jun 22 '13 at 14:05
  • $\begingroup$ You can follow the exact same logic in those cases. Assume that $$0 = \sum_{i=1}^n a_i w_i = \sum_{i=1}^n a_i(u_{i-1}+u_i),$$ use the independency of the $u_i$ to obtain from this a set of equations for the $a_i$, and figure out if it is possible to cook up a non-zero solution to these equations. $\endgroup$ – fuglede Jun 22 '13 at 14:13

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