Let $v={\{v_1,v_2,…,v_k}\}$ Linearly independent
$\mathbb{F} = \mathbb{R}$ or $\mathbb{F}=\mathbb{C}$
Prove that ${\{v_1 + v_2 , v_2+v_3, v_3+v_4,….,v_{k-1}+v_k,v_k+v_1}\}$ is Linearly Independent if and only if $k$ is odd number
$A= S.S$ matrix
$A= \begin{pmatrix}1&1&0&.&.&.&.&0\\ 0&1&1&0&.&.&.&0\\ .&.&&&&&&.\\ .&&.&&&&&.\\ .&&&.&&&&.\\ .&&&&&&&.\\ 0&0&.&.&.&.&1&1\\ 1&0&.&.&.&.&.&1\end{pmatrix}$
to prove this using Determinant I get $|A|=1+(-1)^{k+1}$
so if $k$ is odd number $k+1$ is even and $|A|=2 \rightarrow A$ is Invertible $\rightarrow$ $S$ Linearly Independent
…
but any Idea how to prove this without using Determinant ?
thanks
Best Answer
Consider the alternating sum, $\vec S$ of your terms. If $k$ is even then $\vec S$ is $\vec 0$, so linear independence implies that $k$ is odd.
If $k$ is odd then the sum comes to $2v_1$ so, assuming you are not working over a field of characteristic $2$, $v_1$ is in the span of your elements. Similarly, starting your alternating sum from the $i^{th}$ entry shows that $v_i$ is in your span, and we are done.