${\{v_1 + v_2 , v_2+v_3, v_3+v_4,....,v_{k-1}+v_k,v_k+v_1 }\}$ is Linearly Independent if and only if $k$ is odd number

Question

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

Answer 1

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.

Answer 2

(I used $n$ instead of $k$ below) Begin to row reduce that matrix you have. Use the first row to eliminate the $1$ in the bottom left. The resulting matrix is block upper triangular and the $n-1$-by-$n-1$ block has the same form as your matrix except there is a $-1$ in the bottom left corner. It suffices to show the $n-1$-by-$n-1$ matrix is non-singular. So row reduce, this time adding the first row to the bottom row and getting a block upper triangular matrix where the $n-2$-by-$n-2$ matrix has the same form as your $n$-by-$n$ version. Conclude the result by induction on $n$.

There is also a stronger induction hypothesis that could work, where you add to your assertion that if you take the matrix obtained in the first step of my induction, that this matrix is non-singular if and only if $n$ is even. Then you'll only need to recurse one step in the induction instead of two.