Is there an easy way to compute the characteristic polynomial of a matrix which is "nearly" diagonal? Example: $$ \begin{pmatrix}0 & 0& 0& 1 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0\\ 0 &0 & 0& 0 & 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0\\ 0& 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1\\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\end{pmatrix} $$
it is "nearly" diagonal. If the both entries in the left lower corner would be 0, I just could compute the eigenvalues to be $0$ by $$ det(A-\lambda I)=(-\lambda)^{10} =0 $$