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suppose both A and B are 3X3

I dont have clue on this..any hits?

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- Thread starter yooyo
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- #1

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suppose both A and B are 3X3

I dont have clue on this..any hits?

- #2

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- #3

mathwonk

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how abouit 2Id?

- #4

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what do you mean by 2Id? 2xI?

- #5

matt grime

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- #6

HallsofIvy

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Is the question "Given B find A such that AB= BA"?

- #7

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it is given A find B such that AB=BA..

what is the obvious choice...

what is the obvious choice...

- #8

morphism

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A will commute with itself and all its powers. Also, A will commute with all multiples of Id.

- #9

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it is given A find B such that AB=BA..

what is the obvious choice...

B=A, or B=I.

- #10

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special polynomials in kA, where k is a constant

- #11

matt grime

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Uh? *any* polynomial in A commutes with A (dunno what the k has to do with anything).

- #12

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Uh? *any* polynomial in A commutes with A (dunno what the k has to do with anything).

Hmm... I must have been thinking about something else in particular [which I can't recall now]. Yeah, the k is redundant if I say polynomial.

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Or is there some subtle reason I'm missing why that wouldn't work?

- #14

matt grime

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For any other matrix to work the matrix A must be in a special format. For example, a 2x2 matrix must have the upper right and lower left values equal. It might need to meet another requirement, I'm not too sure.

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matt grime

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matt grime

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I read all of the posts and nobody appeared to answer the posters question to satisfaction. Such anger, I would have expected a better show from someone with such credentials as "homework helper" and "science advisor."

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matt grime

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I read all of the posts and nobody appeared to answer the posters question to satisfaction. Such anger, I would have expected a better show from someone with such credentials as "homework helper" and "science advisor."

Did you read the last post before you made your first post to this thread? The one in which Matt Grimes said

Any polynomial in A will do, as has been said many times in this thread. (I think this is about the 3rd time I've said it.)

How does "any polynomial in A will do" not answer the question? If you don't understand, that is one thing (ask for clarification/amplification, and you will get it). If, on the other hand, you didn't bother to read the prior answers before posting an incorrect answer, that is quite another thing. It tends to piss people off ...

- #22

matt grime

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There may have been easier methods, but seeing as it is early in the school semester and the poster may not be familiar with the other methods, I gave a detailed solution involving elementary terms and operations. After the second post was made, the original poster did not seem to be satisfied, hence my detailed post. In short, don't get so emotional over the internet. If I'm beating a dead horse, so be it.

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hotcommodity, Using your approach, you will have nine equations in nine unknowns, but the space spanned by those nine equations is less than order nine. How exactly do you propose solving those nine equations for the nine unknowns?

On the other hand, Matt Grimes has pointed out one family of easily verifiable solutions, namely [itex]B=\sum_n a_nA^n[/itex].

Edit: There are nine unknowns, not six.

On the other hand, Matt Grimes has pointed out one family of easily verifiable solutions, namely [itex]B=\sum_n a_nA^n[/itex].

Edit: There are nine unknowns, not six.

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matt grime

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There may have been easier methods, but seeing as it is early in the school semester and the poster may not be familiar with the other methods, I gave a detailed solution involving elementary terms and operations.

so you think that someone (anyone) is incapable of noticing that AA=AA, but that thet can solve arbitrary equations in arbitarly unknowns.... Boy, you have the wrong idea....

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I didnt mean for things to get contentious, sorry if I stepped on anyones toes.

- #28

HallsofIvy

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Nah, matt grime is just waking up from hibernation- he's always a little grouchy at that time!

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Oh I see, lol.

- #30

mathwonk

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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.

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