find the eigenvalues of $5×5$ matrix The product of the non-zero eigenvalues of the matrix is ____.


using characteristic equation , it is too lengthy to find eigenvalues of $5×5$ matrix , I'm looking for any short trick to solve this question .
 A: Hint: Note that $(1,0,0,0,1)^T$ and $(0,1,1,1,0)^T$ are eigenvectors by inspection.  What are their corresponding eigenvalues?  (You just need to compute $Av$ for each of the above.)  Now observe that the matrix has only $2$ linearly independent columns, so its kernel has dimension $3$.  Thus $0$ is also an eigenvalue with multiplicity $3$.
A: Let $A$ and $B$ be repectively the following matrices: 
 
Then $A= B +I_5$ where $I_5$ is the identity matrix.
Since B and I commutes then the $eigenvalues(A)=eigenvalues(B) +eigenvalues(I_5)$ (with the same order), that is  $eigenvalue(A)=eigenvalue(B) +1$ (for every eigenvalue of B).
Now commuting the eigenvalues of $B$ is easy since in the first row we have only one nonzero entry.
A: Hint 
Note that $\left(\begin{align}1 \\ 0 \\ 0\\ 0\\ 1\end{align}\right)$ and $\left(\begin{align}1 \\ 0 \\ 0\\ 0\\ -1\end{align}\right)$ are two linearly independent eigenvectors. 
Also note that
$\left(\begin{align}0 \\ 1 \\ 1\\ 1\\ 0\end{align}\right),$ $\left(\begin{align}0 \\ -2 \\ 1\\ 1\\ 0\end{align}\right)$ and $\left(\begin{align}0 \\ 1 \\ 1\\ -2\\ 0\end{align}\right)$ are three linearly independent eigenvectors.
