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Roots of unity of matrices

  1. Dec 5, 2009 #1
    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. jcsd
  3. Dec 5, 2009 #2


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

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

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

    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.
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