Finding Order of A in GL2(2) of \mathbb{Z}_2

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Homework Help Overview

The discussion revolves around finding the order of a specific matrix \( A \) in the general linear group \( GL_2(2) \) over the field \( \mathbb{Z}_2 \). The original poster presents their calculations and seeks a more efficient method for determining the integer \( n \) such that \( A^n = I \), where \( I \) is the identity matrix.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate powers of the matrix \( A \) to find its order and questions whether there is a shorthand method for this process. Some participants challenge the correctness of the original poster's calculations, suggesting that \( A^3 \) does not equal \( I \) and that \( A^n \) may never equal \( I \).

Discussion Status

The discussion is ongoing, with participants exploring the validity of the original poster's calculations and questioning the assumptions made regarding the matrix operations in \( \mathbb{Z}_2 \). Some suggest that trial and error might be the only method available, while others hint at potential patterns or alternative approaches without reaching a consensus.

Contextual Notes

Participants note the significance of working within the field \( \mathbb{Z}_2 \) and the implications this has on matrix calculations. There is an acknowledgment of the limitations of the original poster's approach and the potential for confusion regarding matrix operations in this specific context.

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Homework Statement



The following matrix is an elements of the group GL2(2), that is, the general linear group of 2x2 matrices in [tex]\mathbb{Z}_2[/tex]:

[tex]A = \begin{pmatrix}1 & 1 \\1 & 0\end{pmatrix}[/tex]

Find the order of the element A.

The Attempt at a Solution



I know that the order of A is 3. Because A3=I, where "I" is the identity. I found this by trial and error:

[tex]A^1 = \begin{pmatrix}1 & 1 \\1 & 0\end{pmatrix}^1 \neq \begin{pmatrix}1 & 0 \\0 & 1\end{pmatrix}[/tex]

[tex]A^2 = \begin{pmatrix}1 & 1 \\1 & 0\end{pmatrix} \begin{pmatrix}1 & 1 \\1 & 0\end{pmatrix} \neq \begin{pmatrix}1 & 0 \\0 & 1\end{pmatrix}[/tex]

[tex]A^3 = \begin{pmatrix}1 & 1 \\1 & 0\end{pmatrix} \begin{pmatrix}1 & 1 \\1 & 0\end{pmatrix} \begin{pmatrix}1 & 1 \\1 & 0\end{pmatrix} = \begin{pmatrix}1 & 0 \\0 & 1\end{pmatrix}[/tex].

Here is my question, is there a shorthand method for finding "n" in:

[tex]\begin{pmatrix}1 & 1 \\1 & 0\end{pmatrix}^n = \begin{pmatrix}1 & 0 \\0 & 1\end{pmatrix}[/tex]

Is there any way of solving for n without going through all the suffering matrix multipications above?
 
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Well, your calculation isn't even right. A^3 isn't equal to I. In fact, A^n is never equal to I.
 
Dick said:
Well, your calculation isn't even right. A^3 isn't equal to I. In fact, A^n is never equal to I.

We are working in [tex]\mathbb{Z}_2[/tex]!

[tex]A^3= \begin{pmatrix}3 & 2 \\2 & 1\end{pmatrix}[/tex]

Which is equal to

[tex]I=\begin{pmatrix}1 & 0 \\0 & 1\end{pmatrix}[/tex]

in [tex]\mathbb{Z}_2[/tex]. So, the only way to solve this kind of problem is by trial and error?
 
roam said:
We are working in [tex]\mathbb{Z}_2[/tex]!

[tex]A^3= \begin{pmatrix}3 & 2 \\2 & 1\end{pmatrix}[/tex]

Which is equal to

[tex]I=\begin{pmatrix}1 & 0 \\0 & 1\end{pmatrix}[/tex]

in [tex]\mathbb{Z}_2[/tex]. So, the only way to solve this kind of problem is by trial and error?

Ooops. Missed the Z_2. No, I don't know of any systematic way. Some times it can help to cleverly write the matrix as a sum of others or look for a pattern is small powers. But A^3 is already a small power.
 

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