Qual Problem: When do Matrices Commute?

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    Commute Matrices
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Discussion Overview

The discussion revolves around the conditions under which two matrices commute, specifically examining the case where A + B = AB for nxn matrices A and B. Participants explore the implications of this equation and seek general conditions related to matrix commutation.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant presents a problem involving nxn matrices A and B, stating that if A + B = AB, then it follows that AB = BA.
  • Another participant suggests considering the expression (A-I)(B-I) as a hint to explore the problem further.
  • A participant demonstrates that (A-I)(B-I) leads to the conclusion that BA - AB = 0, thus supporting the claim that AB = BA.
  • Questions are raised about general conditions for matrices to commute, with one participant asking what is necessary and sufficient for AB = BA beyond the given condition.
  • A participant reflects on their approach to rewriting the original equation and explores potential factorizations.
  • One participant mentions that two matrices commute if they are simultaneously diagonalizable, while another notes that this condition only holds if both matrices are diagonalizable.

Areas of Agreement / Disagreement

Participants express differing views on the general conditions for matrix commutation, with some asserting the diagonalizability condition while others highlight its limitations. The discussion remains unresolved regarding the broader implications of matrix commutation.

Contextual Notes

Participants do not fully explore the implications of diagonalizability or other potential conditions for commutation, leaving these aspects open for further discussion.

Who May Find This Useful

Individuals preparing for qualifying exams in mathematics or related fields, as well as those interested in linear algebra and matrix theory.

tornado28
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I'm preparing for a qualifying exam and this problem came up on a previous qual:

Let A and B be nxn matrices. Show that if A + B = AB then AB=BA.
 
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Hint: Consider (A-I)(B-I), where I is the nxn identity matrix.
 
Thanks morphism!

(A-I)(B-I) = AB-A-B+I = AB-AB+I=I. Therefore B-I is the inverse of A-I so we have that I=(B-I)(A-I) = BA-A-B+I = BA-AB+I. Thus BA-AB = 0 as needed.

How did you know to write it that way? Also, do you know any good general conditions related to matrices which commute? What is necessary for AB=BA? What (other than A+B=AB) is sufficient for AB=BA?
 
Last edited:
I wrote it that way after fidgeting around with A+B=AB for a while. If you rewrite this as A+B-AB=0 then you might try to factor out A or B to get A(I-B)+B=0 or B(I-A)+A=0. The symmetry inspired me to subtract I from both sides of the first equation to get A(I-B)+B-I=-I <=> A(I-B)-(I-B)=-I <=> (I-A)(I-B)=I.

As for your other questions, I can't think of anything useful off the top of my head.
 
Thanks. I found another thread with information about when matrices commute. Apparently two matrices commute iff they're simultaneously diagonalizable.
 
tornado28 said:
Thanks. I found another thread with information about when matrices commute. Apparently two matrices commute iff they're simultaneously diagonalizable.
That's only true if the two matrices are diagonalizable to begin with! :)
 

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