How to make a matrix B such that AB=BA

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Discussion Overview

The discussion revolves around finding a matrix B such that the product of matrices A and B commutes, specifically that AB = BA. The context includes theoretical exploration and mathematical reasoning regarding the properties of matrices and their commutativity.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants suggest writing out the matrices explicitly and setting the products equal to find conditions on their entries.
  • One participant proposes using B = 2Id, while another questions the meaning of this suggestion.
  • It is noted that A will commute with itself, its powers, and all multiples of the identity matrix.
  • Some participants mention that any polynomial in A will commute with A, though there is some confusion about the relevance of constants in this context.
  • Another viewpoint suggests that any B sharing all of A's eigenvectors would also work, but this is challenged by a participant who argues that this does not hold without a basis of eigenvectors.
  • There are claims that B could be A^-1 or other specific forms, but these depend on A being invertible or having certain properties.
  • One participant describes a method involving setting up a system of linear equations to find B, while another criticizes this approach as overly tedious.
  • Some participants express frustration over perceived lack of attention to previous posts and the clarity of responses.
  • Several participants mention multiple potential forms for B, including B = 0, B = I, and B = A.
  • A later reply introduces the idea of using a unitary matrix or a specific numerical approach to find B, suggesting a variety of methods exist.

Areas of Agreement / Disagreement

Participants express a range of views on potential forms for B, with no consensus on a single solution. There are disagreements regarding the sufficiency of certain approaches and the clarity of previous responses.

Contextual Notes

Some discussions highlight the need for specific properties of matrix A (such as being invertible) for certain proposed solutions to hold. There are also mentions of the complexity involved in solving the equations derived from the commutation condition.

Who May Find This Useful

Readers interested in linear algebra, matrix theory, or those seeking to understand matrix commutativity may find this discussion relevant.

  • #31
B = 0
B = I
B = A
B = A^-1

There are more I suppose
 
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  • #32
mathwonk said:
enough already! this is a boring question!

here is a more interesting one: prove that a holomorphic map of a riemann surface to itself that induces the identity on homology is the identity map.

Is our surface compact?
 
  • #33
I have a Idea How to Find B that AB=BA!
 
  • #34
Without knowing A?
 
  • #35
hotcommodity already gave the answer, here is another possibility... take a unitary matrix, and let B = A^T plug it in, done !

but if you are desperately looking for numerical values, you can try the following MATLAB code for almost commuting (!?) matrices,

A = rand(5);
B = lyap(A,-A,1e-15*eye(5));
A*B-B*A

I have to perturb the Sylvester equation(since because otherwise it will give the trivial answer B=0. Or you can come up with a automated code that forms the \hat{A}b=0 equation where b is the vectorized entries of B. It is not hard it is just tedious. It will be as

\left\{(I\otimes A) + (-A^T\otimes I)\right\} b= 0

etc... But numerically you should still perturb that and this is not the best way in terms of numerical stability. My favorite will be semidefinite programming... Dig in if you like...

Conclusion, a lot of choices, so pick one, or find the conditions and teach me those.
 
Last edited:
  • #36
trambolin said:
hotcommodity already gave the answer, here is another possibility...

I didn't give the answer (not the first one anyways).

I don't want to get in trouble again, haha :wink:
 
  • #37
I stand by matt_grime. It's frustrating although I'm probably that person from time to time. :smile:

matt_wonk, that's a boring question. :rolleyes: Um... :confused:
 

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