What is the necessary condition for matrix commutation?

Click For Summary

Discussion Overview

The discussion centers on identifying the necessary condition for two arbitrary matrices, A and B, to commute, specifically when the condition is expressed as AB = BA. The scope includes theoretical aspects of matrix algebra and conditions under which matrices commute, particularly in the context of different scalar fields.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant notes that a necessary condition for matrices to commute is that they must both be square.
  • Another participant suggests that matrices must be simultaneously triangularizable as a necessary condition for commuting.
  • A question is raised about the implications when the scalar field is not algebraically closed, specifically regarding real matrices and their commutation properties.
  • One participant proposes that if two matrices are both polynomials in the same matrix, they will commute, indicating a potential necessary condition.

Areas of Agreement / Disagreement

Participants express differing views on what constitutes a necessary condition for matrix commutation, with no consensus reached on a singular necessary condition applicable to all cases.

Contextual Notes

The discussion highlights limitations related to the types of matrices considered (e.g., diagonalizable versus arbitrary) and the implications of the scalar field being algebraically closed or not.

fairy._.queen
Messages
47
Reaction score
0
Hi all!

I was wondering what the necessary condition is for two arbitrary matrices, say A and B, to commute: AB = BA.

I know of several sufficient conditions (e.g. that A, B be diagonal, that they are symmetric and their product is symmetric etc), but I can't think of a necessary one.

Thanks in advance!
 
Physics news on Phys.org
Ok, so it seems the condition (quite sensible actually!) is that they must both be square and simultaneously triangularisable. Thanks a lot!
 
What happens, though, when the matrix scalar field is not algebraically closed? I'm happy with the fact that, in this case, if the two matrices are diagonalisable and commute then they are simultaneously diagonalisable, but what is a necessary condition for arbitrary, say, real matrices to commute (even when they can't be diagonalised)?

Thanks!
 
you could just pretend that the scalar field is the complex numbers, and see if you can make the matrices simultaneously triangular under the complex numbers. If they are not simultaneously triangular under the complex numbers, they will not commute. And if they are simultaneously triangularizable under the complex numbers, then they do commute.
 
Ok, it makes sense. Thanks a lot!
 
fairy._.queen said:
I was wondering what the necessary condition is for two arbitrary matrices, say A and B, to commute: AB = BA.

This remark is from a bound set of notes I found in a used book store: "If two matrices are both polynomials in the same matrix C, then they commute. As we shall see later, this is essentially the only way in which we can have commuting matrices."
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 26 ·
Replies
26
Views
6K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
540