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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.)

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  • $\begingroup$ Did my answer below help? $\endgroup$
    – Fred Hucht
    Commented Jan 3, 2023 at 22:30

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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.

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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.

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  • $\begingroup$ It's true that generically there is nothing to do, but still one would need an algorithm to solve a specific instance of the problem. Matrices that appear in applications may be far from the generic case. $\endgroup$
    – Federico Poloni
    Commented Jul 2, 2022 at 9:04
  • $\begingroup$ Also it would be nice to figure out if the desired value exists at all. It is easy to construct a matrix (say $\begin{bmatrix} 1&\theta\\-\theta & 1\end{bmatrix}$) where the answer is unique ($\theta=0$), but somehow I fail to see an example where there is no such $\theta$ at all... $\endgroup$
    – fedja
    Commented Jul 4, 2022 at 3:23
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    $\begingroup$ a trivial example without any real eigenvalue would be $\begin{bmatrix} \theta&1\\-1&\theta\end{bmatrix}$, I don't have a nontrivial example. $\endgroup$
    – Carlo Beenakker
    Commented Jul 4, 2022 at 8:08

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