Question regarding linearly independent set 
Let $n$ be an integer, $n\geq3$ 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 $$v_i=u_i+u_{i+1} \;\;\ \text{and}\;\; w_i=u_{i-1}+u_i$$ for $i=1,2,...,n$. Then
a) $v_1,v_2,...,v_n$ is linearly independent if $n=2010$
b) $v_1,v_2,...,v_n$ is linearly independent if $n=2011$
c) $w_1,w_2,...,w_n$ is linearly independent if $n=2010$
d) $w_1,w_2,...,w_n$ is linearly independent if $n=2011$

My attempt:
a) $\{v_1,v_2,...,v_n\}=\{u_1+u_2, u_2+u_3,u_3+u_4,...,u_{2010}+u_{2011}\}$
$u_1+u_2=(u_2+u_3)-(u_3+u_4)+\cdots-(u_{2009}+u_{2010})+(u_{2010}+u_1)$ (subtract all odd terms)
So a) is false
b) $\{v_1,v_2,...,v_n\}=\{u_1+u_2, u_2+u_3,u_3+u_4,...,u_{2011}+u_{2012}\}$
$\alpha_1(u_1+u_2)+\alpha_2(u_2+u_3)+\cdots+\alpha_{2011} (u_{2011}+u_1)=0$ implies
$(\alpha_1+\alpha_{2011})u_1+(\alpha_1+\alpha_2)u_2+\cdots+(\alpha_{2010}+\alpha_{2011}) \;u_{2011}=0$
From this , how can I conclude all coefficients are zero ?
What about others? Any help ?
 A: For part b. 
We have $$\alpha_1 + \alpha_{2011} = 0$$
$$\alpha_1 + \alpha_2 = 0$$
$$\vdots$$
$$\alpha_{2010}+\alpha_{2011}=0$$
Let's write this in matrix form.
$$\begin{bmatrix} 1 & 0 & \ldots & \ldots & 0 & 1 \\ 1 & 1 & 0 & \ldots & \ldots  & 0 \\ 0 & 1 & 1 & 0 & \ldots & 0 \\ \vdots & \vdots &  \vdots & \vdots & \vdots &  \vdots \\ & 0 & 0  & \ldots  & 1 & 1\end{bmatrix} \begin{bmatrix} \alpha_1 \\ \vdots \\ \alpha_{2011}\end{bmatrix}=0$$
We can compute the determinant of the coefficient matrix by expanding along the first row which is equal to 
\begin{align}&(-1)^{1+1}\det\left(\begin{bmatrix}   1 & 0 & \ldots & \ldots  & 0 \\   1 & 1 & 0 & \ldots & 0 \\   \vdots &  \vdots & \vdots & \vdots &  \vdots \\ 0 & 0  & \ldots  & 1 & 1\end{bmatrix}\right) + (-1)^{2011+1} \det\left(\begin{bmatrix}   1 & 1 & \ldots & \ldots  & 0 \\   0 & 1 & 1 & \ldots & 0 \\   \vdots &  \vdots & \vdots & \vdots &  \vdots \\ 0 & 0  & \ldots  & 0 & 1\end{bmatrix}\right) \\&= 1 + 1=2 \neq 0\end{align}
Hence we can conclude that $\alpha_i = 0, \forall i \in \{ 1, \ldots, 2011\}$.
Note that the two determinants can be evaluated easily as they are both triangular matrices with $1$ on the diagonals.
