Let V be a vector space with basis $e_1,...,e_n$. Is $e_1 + e_2,e_2 + e_3,e_3 + e_4,...,e_{n−1} + e_n,e_n + e_1$ a basis of this space? So far I have figured out that if you write out zero as the sum of arbitrary coefficients you can deduce that all coefficients have the same absolute value. Further if n is even then $a_i={(-1)}^{i-1}*a_1$, and if n is odd then all coefficients must be 0. Thus if n is even then it is not a basis but if it is odd then it is? What would the truth value of the general case be? My first guess would be false since there is a counterexample but the fact it is true makes me unsure.
 A: Let $v_k:=e_k+e_{k+1}$ for $k=1,\ldots ,n$, where we write $e_{n+1}:=e_1$.
Observe that if $n=3$ then
$v_1-v_2+v_3=2e_3$,  but if $n=4$ then $v_1-v_2+v_3-v_4=0$.
More generally, for every $n>1$,  adding the $v_k$ with alternating signs gives a telescoping sum:
$$\sum_{k=1}^n (-1)^{k-1} v_k=e_1+(-1)^{n-1}e_1 \,. \tag{*}$$
If $n$ is even, then $(*)$ shows that $v_1,\ldots,v_n$ are dependent.
If $n$ is odd, then $(*)$ implies that
$e_1 \in \text{span}\{v_1,\ldots,v_n\}$, and we then deduce inductively that
$e_k \in \text{span}\{v_1,\ldots,v_n\}$ for all $k=1,\ldots,n$. Thus for odd $n$, the vectors $\{v_1,\ldots,v_n\}$ do form a basis of $V$.
A: If $n$ is even, this is never a basis. The easiest case is $n=2$ when you just have two same vectors. If $n=2k$, consider:
\begin{equation*}
1(e_1+e_2)-1(e_2+e_3)+(e_3+e_4)-\cdots +(e_{2k-1}+e_{2k})-(e_{2k}+e_1)=0
\end{equation*}
For example $\{(1,1,0,0),(0,1,1,0),(0,0,1,1),(1,0,0,1)\} \subset \mathbb{R}^4$. We have:
\begin{equation*}
1(1,1,0,0)-(0,1,1,0)+(0,0,1,1)-(1,0,0,1)=0
\end{equation*}
If $n$ is odd, this is always a basis. Let $n=2k+1$ and scalars $a_1, \cdots, a_{2k+1}$ be given such that:
\begin{equation*}
a_1(e_1+e_2)+\cdots+a_{2k+1}(e_{2k+1}+e_1)=0
\end{equation*}
We end up with $a_1+a_{2k+1}=a_1+a_2=a_2+a_3=\cdots=a_{2k}+a_{2k+1}=0$. Easy manipulations show that $a_1=\cdots=a_{2k+1}=0$.
A: Let put aside the special case when $n=1$, if $e_1$ is a base then obviously $2e_1$ is also a base.
The vector independence is maybe easier to visualize in the matrix form. By developing along the first row and with sign alternance rule we get:
$\det(M)=\begin{vmatrix}
1&0&0&0&&0&1\\
1&1&0&0&&0&0\\
0&1&1&0&&0&0\\
0&0&1&1&&0&0\\
&&&\ddots&\ddots&&\\
0&0&0&0&1&1&0\\
0&0&0&0&0&1&1
\end{vmatrix} 
= 1\times \underbrace{\begin{vmatrix}
1&1&0&0\\
0&1&\ddots&0\\
0&0&\ddots&1\\
0&0&0&1
\end{vmatrix}\\}_{=1} 
+ (-1)^{n-1}\underbrace{\begin{vmatrix}
1&0&0&0\\
1&1&0&0\\
0&\ddots&\ddots&0\\
0&0&1&1
\end{vmatrix}\\}_{=1}$
Both minors value is $1$ because both are triangular matrices with all $1$ on the diagonal and all $0$ either in the upper part or the lower part.
Therefore: $$\det(M)=1+(-1)^{n-1}$$
So $M$ is invertible when $n$ odd as $\det(M)=2$ and singular when $n$ even as $\det(M)=0$, and the new set of vectors is a base iff $M$ is invertible.
