# Roots of unity of matrices

1. Dec 5, 2009

### okkvlt

hi. i have recently become very interested in the idea of the nth roots of unity. i have discovered how to calculate them (using eigenvalues), and i find it very fascinating that there are not n many nth roots of unity(unlike scalars).
aparently in the case where the matrix is 2x2, there are n^2 roots of unity

my questions:
given a size=k matrix, find the nth roots of unity. how many roots of unity are there? i want a general formula. is it n^k?
What is the geometrical interperetation of the nth roots of unity of a matrix? the determinants of the nth roots are equal(analogous to the fact that the nth roots of unity of a scalar are points on a circle), but what about the placement of the vectors that compose the matrix? in other words, what is analogous to the fact that the roots of unity of a scalar form a regular polygon?

2. Dec 5, 2009

### Hurkyl

Staff Emeritus
Actually, there are infinitely many square roots! If you solve the system of four quadratic equations in four unknowns, you get the solutions:

$$\left[ \begin{matrix}{\sqrt{1 - xy} & x \\ y & \pm -\sqrt{1 - xy}}\end{matrix} \right]$$

$$\left[ \begin{matrix}{-\sqrt{1 - xy} & x \\ y & \pm \sqrt{1 - xy}}\end{matrix} \right]$$

where x and y can be any numbers at all, such that xy is in [0,1].

Your formula is correct for diagonal matrices: there are exactly n^k n-th roots of unity. This is easy to see, because each diagonal entry can be any n-th root of unity.

But to each such diagonal matrix, there are infinitely many matrices similar to it: for any invertible matrix P, if D is one if your n-th roots of unity, then PDP-1 is another n-th root of unity.