Eigenvalues of the block matrix $C=\begin{bmatrix}−I &-I\\L&0\end{bmatrix}$ Consider the following matrix
$$C=\begin{bmatrix}−I &-I\\L&0\end{bmatrix}$$
where for $L$ we have:
$$L\mathbf{1}=0$$
$$\mathbf{1}^TL=0$$
$$\text{rank}(L)=\dim(L)-1$$
$$L+L^T\geq 0$$
zero is a simple eigenvalue of $L$ and $L+L^T$.
We can show that $C$ has a zero eigenvalue and another one equal to $-1$, what does we can say about the rest of eigenvalues? 
For example, we know that when $L=L^T$, $C$ has only one simple zero and the rest of the eigenvalues have negative real parts.
Numerical examples show that most of the time $C$ has only one simple zero and the rest of the eigenvalues have negative real part, but sometimes there is a pair of eigenvalues with zero real part also appearing. Under what conditions does this pair appear?
 A: We can show $\lambda$ is an eigenvalue of $C$ if and only if $-\lambda(\lambda + 1)$ is an eigenvalue of $L$.
We know $L$ has a one-dimensional nullspace, and setting $-\lambda(\lambda+1) = 0$ shows that $\lambda = 0,-1$ are each eigenvalues of $C$.
Of course that is one way to obtain an eigenvalue of $C$ having zero real part, namely zero.
Is there another possibility?  Let $r$ be an eigenvalue of $L$, perhaps complex.  If $bi$ were a purely imaginary eigenvalue of $C$, we must have:
$$ -bi(bi+1) = r $$
$$ r = b^2 - bi $$
If $L$ is a real matrix, its complex eigenvalues occur in conjugate pairs.  So $\overline{r} = b^2 + bi$ gives rise to another purely imaginary eigenvalue $-bi$ of $C$.
That the characteristic polynomial of $C$ has the form $det(\lambda(\lambda+1)I + L)$ follows from a brief computation:
$$ \lambda I - C = \begin{pmatrix} (\lambda+1)I & I \\ -L & \lambda I \end{pmatrix} $$
Hence our claim, that $\lambda$ is an eigenvalue of $C$ if and only if $\lambda(\lambda+1)$ is an eigenvalue of $-L$.
