Is there an easy way to tell if all eigenvalues of a unitary or self-adjoint matrix only have eigenvalues of multiplicity two?

Question

I am interested in a class of $2n\times 2n$ unitary matrices with complex entries (if you prefer, we can replace "unitary" with "self-adjoint").

I know that all the eigenvalues of matrices in this class have (algebraic) multiplicity one or two. Some very interesting phenomena happens when all the eigenvalues have multiplicity two. Is there a way for me to tell when that happens based on the entries of the matrix, other than computing them by brute force?

Answer 1

Assuming it is feasible to compute the characteristic polynomial, $p(x)$, in your situation, (which can certainly be done in principle if you know the matrix entries) there is a simple strategy. Given the information you already have, all roots have multiplicity two if and only if ${\rm gcd}(p(x),p^{\prime}(x))$ has degree $n.$ Note that this does not require factorization of $p(x),$ it only requires the Euclidean algorithm for polynomials, which is simple to implement computationally (here, $p^{\prime}(x)$ just denotes the derivative of $p(x)$).

Answer 2

I'm assuming that your are working over an algebricaly closed field of characteristic zero (because you seem to be working $\mathbb{C}$). This will also work over $\mathbb{R}$ for self-adjoint matrix.

All the zeros of a polynomial $P$ have multiplicity greater than one if and only if $P$ divide $(P')^2$.

So you can take $P$ to be the characteristic polynomial of your matrix and compute $(P')^2$ in $K[X]/(P)$.

I don't says it is algorithmically more efficient than the previous answer, but it has the advantage of producing for each $n$ a family of $n$ algebraic equations on the coefficient which says if your $2n \times 2n$ matrix satisfy the condition or not...