The following matrix is the result of a special kind of balanced signed graph of order $n$. In the Matrix $n\_1,n\_2,..,n\_k$ are positive integers, which satisfy $\sum
 n\_i=n.$ Prove that this matrix has two zero eigenvalues if and only if $k=6r$ for any positive integer $r$. 


\begin{equation\*}
T\_\lambda=\begin{bmatrix}
-n\_1 & n\_2 & 0 &.&.&0 & n\_k \\ 
n\_1& -n\_2 &n\_3 &0&.&. & 0 \\ 
0& n\_2 & -n\_3 & n\_4 & 0& . & 0 \\ 
 .&0 & n\_3 &-n\_4 &n\_5&0&. \\ 
 .&. &. & . &.&.&. \\ 
.&. &. &. &.&.&. \\
0& 0 &. &. & n\_{k-2}& -n\_{k-1}& n\_{k} \\
n\_1 & 0 & .&.& 0& n\_{k-1}& -n\_k
\end{bmatrix}
\end{equation\*}


I have calculated row reduced echelon forms of the above matrix by the Mathematica. It has two zero eigenvalues for $k = 6r$. I would like to have an analytic proof.


Thank you for your help in advance.



Eigenvalues of cyclic tridiagonal matrix
This seems rather straightforward (and not research level). You have a bunch of 3-term equations $x\_1=x\_k+x\_2$, $x\_2=x\_1+x\_3$, etc., so clearly everything is determined by $x\_1$ and $x\_2$, and you only need to check if you get any relation between them when you "close" the cycle. (Accidentally, the roots of the characteristic polynomial of the corresponding recurrence relation are precisely the 6-th roots of unity.) In more detail, you have
$$
x\_1=\alpha+\beta,\qquad x\_2=\alpha\epsilon\_1+\beta\epsilon\_2,
$$
and
$$
x\_1=x\_{k+1}=\alpha\epsilon\_1^k+\beta\epsilon\_2^k,\qquad
x\_2=x\_{k+2}=\alpha\epsilon\_1^{k+1}+\beta\epsilon\_2^{k+1}
$$
for some $\alpha,\beta$ and $\epsilon\_i$ the primitive $6$-th roots of unity. Eliminate $x\_1$, $x\_2$ and check how many solutions the system in $\alpha,\beta$ has.


