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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

302.0. "Roots of unity / matrices" by TOOLS::STAN () Wed Jun 05 1985 23:29

Let K be the set of 2 by 2 matrices with rational entries.
Let I be the 2 by 2 identity matrix.

For which positive integers, n, does I have an nth root
(other than I itself) in K?

For example,

		-1   0
             (          )
		 0  -1

is a square root of I.  I believe I can show that I has no non-trivial
cube roots in K.  Fourth roots are easy.  What about 5th roots?

Note that if we do not restrict ourselves to rational entries, then
we can always find an nth root, namely

		cos t    sin t
           (			)
	      - sin t    cos t

where t = 2 pi / n.
T.RTitleUserPersonal
Name
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302.1R2ME2::STANThu Jun 06 1985 18:2813
I was wrong about cube roots.

		0   -1
	     (		)
		1   -1

is a cube root of I.  Also

		1   -1
	     (		)
		1    0

is a 6th root of I.  Still not sure if there are any 5th roots of I.
302.2TURTLE::GILBERTThu Jun 06 1985 21:454
In fact, any 2x2 matrix with determinant 1 and trace -1 is a cube root of
the identity matrix.

(the trace of a matrix is the sum of the entries on the major diagonal).
302.3ADVAX::J_ROTHFri Jun 07 1985 06:1725
Try viewing your matrix, A, as representing a linear transformation on a
two dimensional vector space over the reals.

For A**n to equal I, an eigenvalue of A must be the n'th root of unity.

Representing A as an ABCD matrix,

	    |  a   b  |
	A = |         |
	    |  c   d  |

The constraints on its elements are:

	     2	  2	      2		  2
4*cos(2*pi/n)  = a + 2*a*d + d   = (a + d)

	     2			       2		   2
4*sin(2*pi/n)  = 4*(a*d-b*c) - (a + d)   = 4*Det(A) - Tr(A)

1 = (a*d - b*c)

These are clearly satisfied in the examples given.  Since the determinant
of A is the volume measure of the transformation it must equal 1.

- Jim
302.4ADVAX::J_ROTHFri Jun 07 1985 07:0015
Actually, my constraints in .-1 are not as concise as they could be;
one should think before typing reply...

I should have said that:

	Det(A) = (ad - bc) = 1

	Tr(A)   (a + d)
	----- = ------- = cos(2*pi/n)
	  2	   2


Looks like you're out of luck for n=5.

- Jim
302.5ADVAX::J_ROTHSat Jun 08 1985 14:4255
Here's another way of looking at this problem.

Consider the group of Moebius (bilinear) transformations on the
complex plane:

		az + b
	w(z) = --------
		cz + d

We can identify composition of these transformations with multiplication
of the matrices:

	    | a  b |
	A = |      |
	    | c  d |

If Det(A) vanishes, then w(z) reduces to a constant, and is thus non
invertable, as A would be.

Now the problem is to look for periodic sequences of bilinear transformations,
and we can generalize A to be complex.

I believe that its known this is only true for n = 1, 2, 3, 4, and 6
when the elements of A are rational.  This is related to the fact that
the characteristic equation of A is a quadratic, and only the numbers
n = 2, 3, 4, and 6 have 2 or fewer relatively prime integers less than n.
(Euler's function, phi(n) .le. 2). [In my prior note, I should have
referred to 'primitive root of unity of order n' for the eigenvalue of A,
but the idea was right.]

If you stereographically project the complex plane onto the Riemann sphere,
perform some rigid motion of the sphere, and project back I think this is
equivalent to a bilinear transformation; it might be possible to identify
symmetry groups of convex polyhedra on the sphere with periodic sequences
of these transformations but that's just a guess.

Bilinear transformations are familiar to me from microwave and RF filter
design.  The impedance or admittance of a network is related to its
scattering parameters via a bilinear transform.  If the network is
lossless, the S matrix is unitary; this is one key in the synthesis
of a network transfer function on an insertion loss basis.  You can
map periodic structures (transmission lines, sampled digital filters) onto
continuous filters via a bilinear transform (the tan(theta/2) substitution)
and adapt classical filter designs to these areas.  This is where the
Smith chart comes from.  Since circles map to circles its a powerful
design aid since you can plot noise circles and stability circles on
the Smith chart and visualize the boundries of behaviour of an amplifier.
There are network analyzers that directly plot S parameters for you.

Well, so much for my so called 'knowledge' of mathematics...

Check out Schwerdtfeger's book "The geometry of complex numbers" for
more goodies related to this area (one of those inexpensive Dover reprints).

- Jim
302.6R2ME2::STANSat Jun 08 1985 15:443
The transformation is also a unimodular transformation and the
problem is equivalent to asking if there is an affinely regular
n-gon with vertices at rational coordinates in the plane.
302.7ADVAX::J_ROTHThu Jun 13 1985 12:506
I suspected that's what you were in fact up to.  Since the trace and
determinant are invariant under similarity transformation, you couldn't
rotate the n-gon in the plane and somehow increase the number of cases
where it would work, i think.

- Jim