I have two matrices A and C from GL(r,Zn).How can I assert that they wont commute.That is how can I ensure the AC<>CA(AC not equals CA). thanks in advance...

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- Aug 16th 2010, 10:41 PM #1

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- Aug 17th 2010, 01:14 AM #2
You say you "have" two matrices. Does that mean you have the exact matrix representation of these two matrices? If so, why not just compute AC and CA, and see if they're different? Some matrices in your group will commute, and others will not. The identity, for example, commutes with all members of the group!

- Aug 17th 2010, 02:10 AM #3

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- Aug 17th 2010, 02:23 AM #4

- Aug 17th 2010, 02:27 AM #5

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- Aug 17th 2010, 03:10 AM #6
Well, you could try doing this: let

$\displaystyle A=\begin{bmatrix}a_{1} &a_{2}\\ a_{3} &a_{4}\end{bmatrix}$ and

$\displaystyle C=\begin{bmatrix}c_{1} &c_{2}\\ c_{3} &c_{4}\end{bmatrix}.$

Then

$\displaystyle AC=\begin{bmatrix}a_{1}c_{1}+a_{2}c_{3} &a_{1}c_{2}+a_{2}c_{4}\\ a_{3}c_{1}+a_{4}c_{3} &a_{3}c_{2}+a_{4}c_{4}\end{bmatrix}.$

Similarly,

$\displaystyle CA=\begin{bmatrix}a_{1}c_{1}+a_{3}c_{2} &a_{2}c_{1}+a_{4}c_{2}\\ a_{1}c_{3}+a_{3}c_{4} &a_{2}c_{3}+a_{4}c_{4}\end{bmatrix}.$

Therefore, the commutator $\displaystyle [A,C]\equiv AC-CA$ we compute as follows:

$\displaystyle [A,C]=\begin{bmatrix}a_{2}c_{3}-a_{3}c_{2} &a_{1}c_{2}+a_{2}c_{4}-a_{2}c_{1}-a_{4}c_{2}\\ a_{3}c_{1}+a_{4}c_{3}-a_{1}c_{3}-a_{3}c_{4} &a_{3}c_{2}-a_{2}c_{3}\end{bmatrix}.$

For the two matrices to commute, the commutator must be equal to zero. Interesting point: given either $\displaystyle A$ or $\displaystyle C$, you can view the equation $\displaystyle [A,C]=0$ as a system of four linear equations in four unknowns for the unknown matrix. You can then characterize, to some extent, all the matrices that will commute with a given matrix. Note: since the zero matrix and the identity matrix both commute with all matrices, there will always be infinitely many matrices that commute with a given matrix (although, of course, since you're in $\displaystyle Z_{n}$, that will change things a bit. You might have a finite number of commuting matrices in this case.)

That's about as far as I can go.

- Aug 17th 2010, 09:42 PM #7

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- Aug 18th 2010, 04:04 AM #8

- Aug 18th 2010, 04:38 AM #9
So, based on Swlabr's comment, that'll give you two more equations (nonlinear this time, unfortunately) to help characterize your matrices. So you're up to six equations (2 of which are nonlinear, 4 linear) in 8 unknowns.

[EDIT] Actually, I'm wrong. Determinant nonzero gives you two nonlinear inequalities. That's just a check condition at the end, they won't help you solve the system.

- Aug 19th 2010, 09:25 PM #10

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- Aug 20th 2010, 12:10 AM #11

- Aug 20th 2010, 10:43 PM #12

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- Aug 22nd 2010, 11:49 PM #13