# Finding characteristic polynomial of adjacency matrix

Short question im having a tad difficulty with.

I'm trying to find the characteristic polynomial of a graph that is just a circle with n vertices and n edges.

I think the adjacency matrix should look something like this:

$\begin{pmatrix} 0 & 1 & 0 & 0 & \cdots & 1 \\[2mm] 1 & 0 & 1 & 0 & \cdots & 0 \\[2mm] 0 & 1 & 0 & 1 & \cdots & 0 \\ 0 & 0 & 1 & 0 & \ddots & \vdots\\ \vdots & \vdots & \vdots& \ddots& \ddots& 1 \\[2mm] 1 & 0 & 0 & \cdots& 1 & 0\end{pmatrix}$

How do I find the characteristic polynomial of this matrix? The determinant is very difficult to calculate.

• An idea: did you try induction? (Disclaimer: I've no idea whether it'd help but I think it's worth a shot) Jan 6 '14 at 12:28
• Is the last column $(1,0,\cdots,0)$ or is it $(1,0,\cdots,0,1,0)$?
– user114628
Jan 6 '14 at 12:31
• since vertix $n$ is a neighbor to $1$ and $n-1$ the last column is $(1,0,0,\cdots,0,1,0)$ Jan 6 '14 at 12:32
• And yes I just tried induction...maybe there is a way but i didn't get anywhere. I cant see a correlation between case $n=k$ and $n=k+1$ Jan 6 '14 at 12:33

• In his case the eigenvalues are given by $2 \cos \frac{2k\pi}{n}$, $k=1,2,\cdots,n$
• @Kuai: No you are wrong. The difference between the adjacency matrices of a linear (path) and of a cycle graph is only the entries at the extreme bottom-left and top-right, which are $1$ for the cycle, but $0$ for the path graph. The diagonal entries are $0$ in both cases. Oria Gruber, your matrix is correct. Jan 6 '14 at 15:40