Find the eigenvalues of a 5x5 (symmetric) matrix containing a null 4x4 matrix Find the eigenvalues of
$$A=\begin{bmatrix}
    0 & 1 & 1 & 1 &1 \\
    1 & 0& 0 & 0& 0\\
    1 & 0& 0 & 0& 0\\
    1 & 0& 0 & 0& 0\\
    1 & 0& 0 & 0& 0
\end{bmatrix}$$
It doesn't appear to be a block matrix. I can't find a way other than explicitly calculating the determinant of $\det (A-\lambda I)$. Is there something else i'm missing?

The matrix has rank $2$, so the nullspace has dimension $3$. That means $\lambda=0$ is an eigenvalue with multiplicity at least $3$.
Since the matrix is symmetric, it is diagonalizable, so it cannot be that all eigenvalues are equal to $0$.
(Alternatively, since $A^2\neq 0$ (because $A^2(0,1,0,0,0) = A(1,0,0,0,0)\neq (0,0,0,0,0)$) then the characteristic polynomial cannot be $-t^5$ (the Jordan form would have at least three blocks, so the largest block has size at most $2$, and then $A^2$ would be zero).
The trace is $0$, so that means that the other two eigenvalues are of the form $\lambda$ and $-\lambda$ for some $\lambda\neq 0$.
If $(a,b,c,d,e)$ is an eigenvector of $\lambda$, with $\lambda\neq 0$, then $\lambda a=b+c+d+e$, and $a=\lambda b=\lambda c=\lambda d=\lambda e$. So $b=c=d=e = \frac{a}{\lambda}$, which gives $\lambda^2=4$. Thus, if there are any nonzero eigenvalues, then they should be $\lambda=2$ and $\lambda=-2$. So the eigenvalues are $0$ (with multiplicity $3$), $2$, and $-2$.
