- #1

- 82

- 0

this is what i got so far.

AB=BA

AB=B^(2)A^(2)

AB=(BA)^(2)

this is where I get stuck.

Do A and B have inverses? if so, why?

should I be thinking about inverses or is there another way of approaching this problem?

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- Thread starter eyehategod
- Start date

- #1

- 82

- 0

this is what i got so far.

AB=BA

AB=B^(2)A^(2)

AB=(BA)^(2)

this is where I get stuck.

Do A and B have inverses? if so, why?

should I be thinking about inverses or is there another way of approaching this problem?

- #2

- 229

- 0

this is what i got so far.

AB=BA

AB=B^(2)A^(2)

AB=(BA)^(2)

this is where I get stuck.

Do A and B have inverses? if so, why?

should I be thinking about inverses or is there another way of approaching this problem?

(AB)^2 = ABAB = AABB = A^2B^2 = AB

- #3

- 82

- 0

can you just switch the B and A from ABAB to get AABB?

- #4

- 631

- 0

ABAB = A(BA)B = A(AB)B = AABB. Is that OK ?

Last edited:

- #5

- 82

- 0

- #6

learningphysics

Homework Helper

- 4,099

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It is given in the question that AB = BA... so it's ok to switch them.

- #7

- 1,074

- 1

YOU said in the first post AB=BA, IF that is true then you can switch the order like that.

- #8

matt grime

Science Advisor

Homework Helper

- 9,420

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As has been pointed out THEY COMMUTE! But that isn't why I post. I want to point out that only in the trivial case can an idempotent be invertible.

Last edited:

- #9

- 82

- 0

so if i start the proof off with AB=BA then I can use AB=BA later on in the proof I started off with in the firszt place?

Last edited:

- #10

learningphysics

Homework Helper

- 4,099

- 6

so if i start the proof off with AB=BA then I can use AB=BA later on in the proof I started off with in the firszt place?

Yes. you're given AB = BA is true... so you can use that anywhere in your proof...

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