Conditions under which a $2\times2$ block matrix has complex eigenvalues Consider a matrix $A$ in $\mathbb{R}^2$:
\begin{equation}
A=\begin{bmatrix}
0 & -c \\
1 & -b
\end{bmatrix}
\end{equation}
Then it can be shown that the matrix has complex eigenvalues if $b^2-4c < 0$. Can a similar relation be derived also for the elements of a similarly shaped 2x2 block matrix, ie:
\begin{equation}
M=\begin{bmatrix}
0_n & -C \\
I_n & -B
\end{bmatrix}
\end{equation}
where $B, C$ are matrices in $\mathbb{R}^n$, such that $\lambda_M$ are complex?

First we note that the elements $(2,1)$ and $(2,2)$ of matrix $M$ commute. Then we consider:
\begin{equation}
(\lambda I_{2n}-M)v = 0
\end{equation}
which is the eigenvalue problem of $M$. The equation admits a solution $v \neq 0$ when $|\lambda I_{2n}-M| = 0$, so we resort to computing the determinant of the characteristic matrix:
\begin{equation}
\begin{bmatrix}
\lambda I_n & -C \\
I_n & \lambda I_n + B \\
\end{bmatrix}
\end{equation}
We assume that $(1,1)$ is invertible, i.e., that $0$ is not an eigenvalue of $M$. Then since $I_n$ commutes with any matrix, we can write:
\begin{equation}
|\lambda I_{2n} - M| = |\lambda ( \lambda I_n + B) + C | = |\lambda^2 I_n + \lambda B + C| = 0
\end{equation}
This is a Quadratic Eigenvalue Problem and has been intensively studied. Particularly, in (Tisseur, 2001), properties of the solutions $\lambda$ have been given in terms of the coefficient matrices $M, C, K$ of $\lambda^2 M + \lambda C + K = 0$. When the three matrices are Hermitian, then the eigenvalues are some solutions of:
\begin{equation}
\lambda = \frac{-c(x) \pm \sqrt{c(x)^2 - 4k(x)m(x)}}{2m(x)}
\end{equation}
with $m(x) = x^*Mx,c(x) = x^*Cx, k(x) = x^*Kx$. Then in this case it can be seen that a sufficient, but not necessary condition for $\lambda$ to be complex is that $c(x)^2 < 4k(x)m(x)$ $\forall x$, which in our case reduces to
\begin{equation}b(x)^2 < 4c(x) \text{  }  \forall x\end{equation}
with $b(x) = x^* B x, c(x) = x^* C x$ (pay attention to the different meaning of $c(x)$ in the problem of Tisseur and the answer's problem).
If $B, C$ are Hermitian (which, being them real, implies symmetry), then $\lambda_{C,n} \leq c(x) \leq \lambda_{C,1}$ and $\min(|\lambda_{B}|)^2 \leq b(x)^2 \leq \max(|\lambda_{B}|)^2$, so a (very) broad sufficient condition is that:
\begin{equation}
 \min(\lambda_C) > \bigg(\frac{\max(|\lambda_{B}|)}{2}\bigg)^2
\end{equation}
If they are not Hermitian, then the same condition can be stated in terms of their symmetric parts since $b(x) = x^*Bx = \frac12 x^*(B+B^T)x$ and similarly for $c(x)$.
