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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 ?

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1 Answer 1

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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.

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  • $\begingroup$ Superb! I think using the same determinant technique to decide c) and d) are also true $\endgroup$ Aug 4, 2017 at 7:02
  • $\begingroup$ What about your opinion? $\endgroup$ Aug 4, 2017 at 7:15
  • $\begingroup$ determinant trick would work, in fact $c$ and $d$ should be simpler. $\endgroup$ Aug 4, 2017 at 7:17
  • $\begingroup$ For c), $a_1w_1+a_2w_2+\cdots+a_nw_n=0$ implies $a_1u_1+a_2(u_1+u_2)+a_3(u_2+u_3)+\cdots+a_{2009}(u_{2008}+u_{2009})+a_{2010}(u_{2009}+u_{2010})=0$ implies $(a_1+a_2)u_1+\cdots+a_{2010}u_{2010}=0$. By independence of $u_i$'s, $a_{2010}=0$ and by backward process all coefficients must be zero. Am i right? $\endgroup$ Aug 4, 2017 at 7:28
  • $\begingroup$ Yup, that is correct. $\endgroup$ Aug 4, 2017 at 7:29

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