Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
    
Are you interested in an algorithm to run on a computer or in a theoretical tool? If the former, do you want an exact result (all computations with big integers or exact rationals) or are you fine with an implementation based on floating point numbers? –  Federico Poloni Aug 21 at 9:38
    
I'm interested in a theoretical tool. Sorry for making that point unclear. –  Darren Ong Aug 21 at 13:55

2 Answers 2

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 $p(x)$).

share|improve this answer

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...

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.