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?
 A: 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 $-2$.
A: We have $A=XY$ where
$$
X=\pmatrix{1&0\\ 0&1\\ 0&1\\ 0&1\\ 0&1}
\quad\text{and}\quad Y=\pmatrix{0&1&1&1&1\\ 1&0&0&0&0}.
$$
By Sylvester's secular theorem, $XY$ and $YX$ share the same multi-set of nonzero eigenvalues. The eigenvalues of $YX=\pmatrix{0&4\\ 1&0}$ are $\pm2$. Therefore the eigenvalues of $A$ are $2,-2,0,0,0$.
Alternatively, consider the case where $A$ is a real matrix first. In this case, $A$ is a rank-two real symmetric matrix with trace $0$ and Frobenius norm $\sqrt{8}$. Hence it has precisely two nonzero real eigenvalues $\pm\lambda$, with $2\lambda^2=8$. Therefore $\lambda=\pm2$ and the spectrum of $A$ is $\{2,-2,0,0,0\}$. Since $A$ is an integer matrix with integer eigenvalues, if we view $A$ as a matrix over another field, it will have the same spectrum.
A: We can generalize the problem to finding eigenvalues (and eigenvectors) of the following  $(n+1)\times(n+1)$
Hermitian matrix
$$
M=\left(\begin{array}{cc}
0 & x^{\dagger}\\
x & \mathbf{0}
\end{array}\right),
$$
where
$$
x =\left(\begin{array}{c}
x_1\\
x_2 \\
\vdots \\
x_n
\end{array}\right).
$$
Making the Ansatz
$$
v =\left(\begin{array}{c}
a\\
y
\end{array}\right),
$$
where
$$
y =\left(\begin{array}{c}
y_1\\
y_2 \\
\vdots \\
y_n
\end{array}\right),
$$
the eigenvalue equation
$$ Mv = \lambda v$$
implies
\begin{align}
  \lambda a & = x^\dagger y \\
\lambda y &=  a x. 
\end{align}
One readily finds two eigenvalues given by
$$ \lambda _\pm = \pm \left\Vert x\right\Vert $$
with eigenvectors
$$
v_{\pm}=\left(\begin{array}{c}
1\\
\pm\frac{x}{\left\Vert x\right\Vert }
\end{array}\right).
$$
All other eigenvalues are zero with eigenvectors given by
$$
v_{y}=\left(\begin{array}{c}
0\\
y
\end{array}\right),
$$
with $x^{\dagger}y=0$, and they are $n-1$.
Your problem has $n=4$ and $x_i =1$, $i=1,\ldots,4$.
A: This solution is essentially @Arturo Magidin's, but tweaked slightly. For each eigenvalue $\lambda$, let $a_{\lambda}$ denote the algebraic multiplicity of $\lambda$ and let $E_{\lambda}$ denote the eigenspace associated to $\lambda$.
The matrix $A$ is diagonalizable because $A$ is symmetric; hence $a_{\lambda} = \dim E_{\lambda}$ for each eigenvalue $\lambda$. $A$ clearly has rank 2, which implies that
$$
3 = \dim(\operatorname{Nul} A) = \dim E_{\lambda = 0} = a_{\lambda = 0}.
$$
Thus there are two other nonzero eigenvalues $\lambda_{1}$ and $\lambda_{2}$ (that are a priori not necessarily distinct). Since $\operatorname{Tr} A = 0$, we have $\lambda_{1} + \lambda_{2} = 0$, which implies that $\lambda_{2} = - \lambda_{1}$.
Now follow @Arturo Magidin's answer to determine $\lambda_{1}$ and $\lambda_{2}$.
