Consider the (real) linear system of equations $A\mathbf{x}=\mathbf{c}$ of size $N$ as
$$
\begin{bmatrix} a\_{N}-a\_{2}& a\_{2}& 0 &\dots& 0 & -a\_{N} \\-a\_{1} & a\_{1}-a\_{3}&a\_{3}& 0 & \dots&0\\
0 & -a\_{2} & a\_{2}-a\_{4}&a\_{4}&0 & \dots \\\vdots&\ddots&\ddots&\ddots&\ddots\\ a\_1& 0& \dots& 0&-a\_{N-1}&a\_{N-1}-a\_1 \end{bmatrix}\begin{bmatrix}
x\_1\\x\_2\\x\_3\\\vdots\\x\_N\end{bmatrix}=\begin{bmatrix}
c\\c\\c\\\vdots\\c\end{bmatrix}.
$$
So, the vector $\mathbf{c}$ is constant and $A$ has a periodic structure and its kernel (at least) contains constant vectors (since the sum of columns is zero). 


How can one prove that


1. $rank(A)=N-1$
2. The system is ***inconsistent***, unless $c=0$.


I suppose that these claims are valid. But any counter-example is also very appreciated.



How to show inconsistency of a linear system of equations?
For a counterexample to (1) with all $a\_i$ nonzero, consider the $N=4$ case with $a\_3 = a\_1$, noting that
$$ \left[\matrix{a\_4-a\_2 & a\_2 & 0 & -a\_4\cr
-a\_1 & 0 & a\_1 & 0\cr
0 & -a\_2 & a\_2-a\_4 & a\_4\cr a\_1 & 0 & -a\_1 & 0\cr}\right] \left[\matrix{0 \cr a\_4 \cr 0 \cr a\_2}\right] = 0$$


For a counterexample to (2) with all $a\_i$ nonzero, consider the $N=3$ case with $a\_1=4, a\_2=a\_3=1$:
$$ \left[\matrix{0 & 1 & -1\cr -4 & 3 & 1\cr 4 & -1 & -3\cr}\right] \left[ \matrix{1 \cr 2\cr 0\cr}\right] = \left[\matrix{2\cr 2\cr 2\cr}\right]$$


