How can I compute eigenvalues or characteristic polynomial of this matrix? Please help. \begin{pmatrix}
2na & -a & -a & -a & -a &  -a& -a\\ 
 -a& a+b & 0 & 0 & -b & 0 & 0\\ 
 -a& 0 & a+b & 0 & 0 & -b &0 \\ 
 -a& 0 & 0 & a+b & 0 &  0&-b \\ 
 -a& -b & 0 & 0 & a+b & 0 & 0\\ 
 -a&  0&-b  & 0 & 0 & a+b &0 \\ 
 -a&  0&  0&-b  & 0 & 0 & a+b
\end{pmatrix} 
I represent the matrix when $n=3$. There are block matrices $(a+b)I_3$ and $(-b)I_3$ above. I think the characteristic polynomial of this matrix for any natural number $n$ is calculated efficiently.
The rank is $2n$, so the characteristic polynomial should zero as the constant part. But I cannot calculate the determinant and I can't get zero when I put $t=0$ when I calculate $\text{char}(M)(t)$ for any $n$. Please give me a favor.
 A: Add to row 1 all other rows. Then subtract column 1 from every other column. Your matrix is then similar to
$$
B=\left[\begin{array}{c|c}0&0\\ \hline\ast&C\end{array}\right]
=\left[\begin{array}{c|cc}
0&0&0\\
\hline
\ast&(a+b)I_n+aE&-bI_n+aE\\
\ast&-bI_n+aE&(a+b)I_n+aE
\end{array}\right]
$$
where $E$ denotes the $n\times n$ matrix of ones. Thus the spectrum of your matrix (or $B$) comprises of a zero and the $2n$ eigenvalues of $C$. Since all sub-blocks of $C$ commute, its eigenvalues are the roots of $\det\left([tI_n-(a+b)I_n-aE]^2 - (-bI+aE)^2\right)=0$, which are the two roots of $[t-(a+b)-na]^2 - (na-b)^2=0$ (each of multiplicity $1$) as well as the two roots of $[t-(a+b)]^2 - b^2 = 0$ (each of multiplicity $n-1$).
So, the complete spectrum of the original matrix is
$$
\begin{cases}
0       &\text{(of multiplicity $1$)},\\
(2n+1)a &\text{(of multiplicity $1$)},\\
a+2b    &\text{(of multiplicity $n$)},\\
a       &\text{(of multiplicity $n-1$)}.
\end{cases}
$$
Edit. Knowing the eigenvalues, it is not hard to find the eigenvectors of the original matrix by inspection:
\begin{cases}
\lambda=0:       &v=e_1+\ldots+e_{2n+1},\\
\lambda=(2n+1)a: &v=-2ne_1+(e_2+\ldots+e_{2n+1}),\\
\lambda=a+2b:    &v=e_{1+i}-e_{1+i+n}\quad (i=1,2,\ldots,n),\\
\lambda=a:       &v=\sum_{i=1}^k (e_{1+i}+e_{1+i+n}) - k(e_{k+2}+e_{k+2+n})\quad (k=1,2,\ldots,n-1),
\end{cases}
where the eigenvectors for $\lambda=a$ are constructed in the spirit of "Jagy matrix". These eigenvectors are already mutually orthogonal to each other. So, if you want an orthonormal eigenbasis, just normalise each of them to a unit vector.
