Linear Algebra Theory Question - Heisenberg Group

[s.perkins]

Homework Statement

Let H be the set of all matrices with entries that are integers, with 1s on the main diagonal, and 0s below it.

A = [1 a b; 0 1 c; 0 0 1]

Now, if a set of matrices G contains the identity matrix, contains the inverse matrix of every matrix in G, and is closed under matrix multiplication, it is called a matrix group. The group described prior (H) is called the Heisenberg group. Now find the most general matrix C that belongs to H and commutes with all the elements of H, meaning that CA = AC for every A in H.

Homework Equations

The Attempt at a Solution

I determined that A^2 and A^-1 still belong to H, but I have no idea how to set this up. Any hints would be greatly appreciated. :)

 

[mighty2000]

Hello there,

It seems like you are trying to find the most general matrix C that satisfies the commutativity condition with all elements in the Heisenberg group H. Let's break down the problem into smaller steps.

First, let's consider the commutativity condition for two matrices A and B:

CA = AC

This means that the product of C and A must be equal to the product of A and C. We can expand this to include the elements of A and B:

[1 a b; 0 1 c; 0 0 1] [c1 c2 c3; c4 c5 c6; c7 c8 c9] = [c1 c2 c3; c4 c5 c6; c7 c8 c9] [1 a b; 0 1 c; 0 0 1]

This gives us a system of equations that we can solve to find the values of c1-c9. However, since we are looking for the most general matrix C, we want to find a pattern or a rule that applies to all matrices in H, rather than solving for specific values.

To do this, let's consider the structure of the matrices in H. We have 1s on the main diagonal, and 0s below it. This means that the only non-zero elements in the first row of C must be c1, c2, and c3. Similarly, the only non-zero elements in the second row of C must be c4, c5, and c6. And finally, the only non-zero elements in the third row of C must be c7, c8, and c9.

Now, let's consider the first column of the product CA and AC:

[1 0 0; 0 1 0; 0 0 1] [c1 c2 c3; c4 c5 c6; c7 c8 c9] = [c1 c2 c3; c4 c5 c6; c7 c8 c9] [1 0 0; 0 1 0; 0 0 1]

This gives us the following equations:

c1 = c1
c4 = c4
c7 = c7

This means that the first column of C must have the same values as the first column of A. Similarly, the second and third