# A little problem involving unitary matrices

Assume U is a NxN unitary matrix. Further assume that for all k<n: Tr(U^k)=0. What is the larges possible value for n?

jim mcnamara
Mentor
Definiton question --
You're using what seems to be only reals for unitary matrices, which are complex. Do you mean orthogonal matrix?

matt grime
Homework Helper
Eh? Where does he use anything about reals?

I could mention that I have a very strong hunch that the largest possible value for n is N, and that this value is reached for unitaries of the form $$_{kl} = \delta_{k+1 mod(N),l}$$.

Last edited:
matt grime
Homework Helper
Well, n is certainly bound above by N-1, and that bound can be attained. It is a nice exercise to show that, passing to an algebraic closure as necessary, that for an NxN matrix Tr(X^r)=0 for all r from 1 to N inclusive implies that X is nilpotent - this is becuase these polys are a basis for the symmetric polys in the N eigenvalues (counted with multiplicities) of X, and if they are all zero then so is the product of all the eigenvalues as that is another symmetric poly, which in turn implies one e-value is zero, and by induction all e-values are zero, and X is nilpotent. Since unitary matrices are not nilpotent that puts N as the strict upper bound on n in your question. Certainly 1 is attainable for 2x2 matrices, and it is easy to see that you can get n=N-1 for N prime. I haven't checked your example, but I see no reason not believe you haven't checked it.

Matt you don't happen to have a good reference for what you wrote above? You see, I'm using the fact that Tr(U^k) can't be zero for all k if U is unitary in a text I'm writing, but I do not want to litter the text with details regarding this fact. Any help would be highly appreciated.

matt grime