Finding $\theta$ such that at least one eigenvalue of $A(\theta)$ is real Is there a known method to find a set of $\theta$ such that at least one eigenvalue of $A(\theta)$ is purely real?
Assume $A(\theta)$ is a real square matrix whose elements are linear functions of a real scalar $\theta$. I am interested in a general case (not assuming $A$ is symmetric/positive-definite/etc.)
 A: Let $A(\theta)$ be a real $n\times n$ matrix. If $n$ is odd, there must be at least one real eigenvalue due to complex conjugate pairs, see the argument of @Carlo.
If $n$ is even, you can discuss the discriminant $d(p(\lambda);\theta)$ of the characteristic polynomial $p(\lambda)=\det(A(\theta)-λ 1)$.
If $$(-1)^{n/2} d(p(\lambda);\theta)<0,$$ there are at least two real eigenvalues. At every simple zero of $d(\theta)$, the number of real eigenvalues changes by two. Hence, $(-1)^{n/2} d(p(\lambda);\theta)>0$ is a necessary, but not sufficient condition for no real eigenvalues.
A: Since the eigenvalues of a real matrix $A(\theta)$ come in complex conjugate pairs, an eigenvalue on the real axis with multiplicity 1 cannot move off the real axis when $\theta$ is varied over a small interval. It must first merge with a second real eigenvalue. This means that the set of $\theta$ where $A(\theta)$ has at least one real eigenvalue consists of the union of intervals $[\theta_1,\theta_2]$, $[\theta_3,\theta_4]$, $\ldots$.
Generically, an $n\times n$ real matrix $A$ will have on the order of $\sqrt{n}$ real eigenvalues, so if $n$ is large and you pick a random $\theta$, you are likely to find at least one real eigenvalue. No fine tuning of $\theta$ is needed. This seems the most efficient way to solve the problem.
