Given $B=\{v_1,v_2,...,v_n\}$ basis what is the necessary condition so that $C=\{v_1+v_2,v_2+v_3,...,v_n+v_1\}$ will also be a basis? 
Given $B=\{v_1,v_2,...,v_n\}$, a basis in $\mathbb{R^n}$ what is the necessary condition so that $C=\{v_1+v_2,v_2+v_3,...,v_n+v_1\}$ will also be a basis in $\mathbb{R^n}$?

The answer to this question has to use the fact $\mathbf{(1)}$ that matrix $A_{n \times n}$ of form:
$$
\begin{bmatrix}
1&0&.&.&.&.&1 \\
1&1&0&.&.&.&0\\
0&1&1&.&.&.&.\\
.&.&.&.&.&.&.\\
.&.&.&.&.&.&.\\
0&.&.&0&1&1&0\\
0&.&.&.&0&1&1
\end{bmatrix}
$$
has det$(A)=0$ if $n$ is even while det$(A) \neq0$ if $n$ is odd.
We can see that $A$ can be received from $I_n$ after the operations that are described in $C$. I'm not sure how to prove this for any basis which is not the same as $I_n$. If I could prove that $C$ is row equivalent to $A$ I could use (1) for proof. 
 A: If $\{v_1, \ldots, v_n\}$ spans a vector space $V$, then for any linear transformation $A$, $\{Av_1, \ldots, A v_n \}$ spans the image of $A$; $A$ is invertible iff its image spans $V$, so $\{A v_1, \ldots, A v_n\}$ spans $V$ iff $A$ is invertible. For sets of $n$ vectors in an $n$-dimensional vector space, of course, "spans $V$" and "is a basis of $V$" are equivalent. For your particular matrix $A$, we have $A v_1 = v_1 + v_2$, $A v_2 = v_2 + v_3$, and so forth, for any set of vectors $\{v_1, \ldots, v_n\}$, not just for the standard basis. You should be able to draw conclusions from here.
A: If you want to do everything with determinants, you can reason as follows: if $M$ is an $n\times n$ matrix, then its columns form a basis of $\Bbb R^n$ if and only if $\det(M)\neq0$. Now if $B,C$ are the matrices whose columns describe (in coordinates) the $n$-tuples of vectors of the same name, then by construction $C=B\cdot A$, and so $\det(C)=\det(B)\det(A)$. It is given that $B$ describes a basis, so $\det(B)\neq 0$; then $\det(C)\neq0$ if and only if $\det(A)\neq0$, which apparently you know to be the case if and only if $n$ is odd.
