Re: Matrix Algebra question

On Jun 13, 2008 10:02 PM CT, TCL wrote:

Let L_2 be the 2x2 lower triangular matrix whose
nonzero off diagonal entry is 2, i.e. a11=1, a12=0,
a21=2, a22=1. Let U_2 be its transpose.
I am looking for an easy proof of the following fact:

The group (with matrix multiplication) generated by
{L_2, U_2} is the set of matrices A with a11, a22 odd,
and a21, a12 even, and det(A)=1.

A direct proof seems to be not easy.

Here are some facts that may help (I will denote L_2 and
U_2 as just L and U, respectively).

First, we can show that L (and hence U) has a unit
determinant. For if L = [a_11, 0; 2, a_22], then L^{-2}
= [a_22, 0; -2, a_11]. Thus, det(L) = a_11 a_22 =
det(L^{-1}), so

1 = det(I) = det(L L^{-1}) = det(L) det(L^{-1})

...which implies...

det(L) = det(L^{-1}) = +- 1.

Furthermore, it is not difficult to diagonalize L by
finding the eigenvalues and eigenvectors -- specifically

lambda_1 = a_11 with v_1 = [d, 2], and

lambda_2 = a_22 with v_2 = [0, 1]

...where d = a_11 - a_22. It follows from this that

L = P J P^{-1}

...where J = [a_11, 0; 0, a_22], P = [d, 0; 2, 1],
P^{-1} = [1/d, 0; -2/d, 1]. Note that this implies

L^n = P J^n P^{-1}

...for all positive integers n. Moreover, since U = L^t
we see that

U = L^t = (P J P^{-1})^t = P^t J (P^{-1})^t.

Again, note that this implies that

U^n = P^t J^n (P^{-1})^t

...for all positive integers n.

Maybe we can use these general facts of L^n and U^n to
prove your proposition about G = < L, U >.

Regards,
Kyle Czarnecki
.