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I am stuck with the following problem:

Let $n$ be an integer where $n \geq 3,$ and let $u_1,u_2,.....,u_n$ be n linearly independent elements in a vector space over $\Bbb R$. Set $u_0=0$ and $u_{n+1}=u_1.$ Define $v_i=u_i+u_{i+1}$ and $w_i=u_{i-1}+u_i$ for $i=1,2,3,....,n.$ Then which of the following options are correct?
1. $v_1,v_2,....,v_n$ are linearly independent ,if $n=2010.$
2. $v_1,v_2,....,v_n$ are linearly independent ,if $n=2011.$
3. $w_1,w_2,....,w_n$ are linearly independent ,if $n=2010.$
4. $w_1,w_2,....,w_n$ are linearly independent ,if $n=2011.$

My Attempt: I compute $a_1v_1+a_2v_2+.....+a_nv_n=a_1(u_1+u_2)+a_2(u_2+u_3)+....+a_n(u_n+u_1)=(a_1u_1+...+a_nu_n)+(a_1u_2+...+a_nu_1)=a_1u_2+...+a_nu_1$

[since $u_1,u_2,.....,u_n$ be n linearly independent elements in a vector space over $\Bbb R$ (given),$a_1u_1+...+a_nu_n=0$].Now I do not know which way to progress. Since I am supposed to check $v_1,v_2,....,v_n$ are linearly independent ,my aim is to find the value of $n$ for which $a_1v_1+a_2v_2+.....+a_nv_n=0 \implies a_1=a_2=...=a_n=0$.

Can someone point me in the right direction? Thanks in advance for your time.
EDIT: Now as suggested by @Andreas Caranti
I see that for $n=3, a_1v_1+a_2v_2+a_3v_3=a_1(u_1+u_2)+a_2(u_2+u_3)+a_3(u_3+u_1)=u_1(a_1+a_3)+u_2(a_1+a_2)+u_3(a_2+a_2)=0 \implies a_1=a_2=a_3=0. $

So ${v_1,v_2,v_3}$ are L.I. So, does it mean option $1$ is true ($n=2010$ being a multiple of $3$)?

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2 Answers

Try the small cases, like $n = 3, 4$, and it should be clear which way to go.

Concerning what you have done so far, look at the vectors $$ u_1+u_2, u_2+u_3, \dots, u_n+u_1. $$ Can you see a linear equation vanishing on all of them? Try the two cases $n = 3, 4$ and see whether they are different in this respect.

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Hints. For (1) and (2), let $U=(u_1,\ldots,u_n),\,V=(v_1,\ldots,v_n)$. Then $$ V=U\underbrace{\begin{pmatrix}1&&&&1\\1&1\\&\ddots&\ddots\\&&\ddots&\ddots\\&&&1&1\end{pmatrix}}_{A}. $$ What is $\det A$?

For (3) and (4), what is $w_i-w_{i-1}+w_{i-2}-w_{i-3}+\ldots+(-1)^{i-1}w_1$? Alternatively, let $W=(w_1,\ldots,w_n)$. Then $$ V=W\underbrace{\begin{pmatrix}1&1\\&1&1\\&&\ddots&\ddots\\&&&\ddots&1\\&&&&1\end{pmatrix}}_{B}. $$ What is $\det B$?

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I see that det$A \neq 0$ for $n=3$ and det$A =0$ for $n=4.$ So for odd values of $n,|A| \neq 0.$ Similar case will happen for options $(3)$ and $(4).$ So out of $(1)$ and $(2),$ option $1$ is true.Am I right? –  user52976 Mar 24 '13 at 15:32
    
@user33640 Your "so for odd values of $n$, $|A|\neq0$" seems to me an application of the law of small numbers here. You can compute $\det A$ using Laplace expansion by the first row. As for (3) and (4), I don't understand why you say that "similar case will happen", because my hint for (3) and (4) are different from the hint for (1) and (2). But I have made a small mistake here. See my new edit. –  user1551 Mar 24 '13 at 16:22
    
Thanks for the clarification. As for $(3)$ & $(4),$ I see that for $i=2010,\space -w_1+w_2-.....+w_i=u_{2010}$ and $i=2011,\space w_1-w_2+.....+w_i=u_{2011}.$ Now I can not progress. –  user52976 Mar 24 '13 at 16:39
    
@user33640 Consider $w_i-w_{i-1}+w_{i-2}-w_{i-3}+\ldots+(-1)^{i-1}w_1$ for every $i\in\{1,2,\ldots,n\}$, not just $i=n$. –  user1551 Mar 24 '13 at 16:47
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@user33640 Express $w_1,\ w_2-w_1,\ w_3-w_2+w_1, \ldots$ in terms of $u_i$s only. –  user1551 Mar 24 '13 at 17:21
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