how to make a matrix B such that AB=BA


by yooyo
Tags: abba, matrix
yooyo
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#1
Feb14-07, 10:49 PM
P: 7
B is not the identity matrix or the zero matrix
suppose both A and B are 3X3
I dont have clue on this..any hits?
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d_leet
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#2
Feb14-07, 10:53 PM
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Well write out two 3x3 matrices of variables and calculate AB, and BA set them equal to one another(element by element) and find the conditions on the entries of each matrix that make it true that AB=BA.
mathwonk
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#3
Feb14-07, 11:10 PM
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how abouit 2Id?

yooyo
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#4
Feb14-07, 11:44 PM
P: 7

how to make a matrix B such that AB=BA


what do you mean by 2Id? 2xI?
matt grime
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#5
Feb15-07, 03:35 AM
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There is an obvious choice for B (well infinitely many) that is not Id or 0. What will A always commute with?
HallsofIvy
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#6
Feb15-07, 04:26 AM
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Thanks
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Is the question "Given B find A such that AB= BA"?
yooyo
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#7
Feb15-07, 09:57 PM
P: 7
it is given A find B such that AB=BA..
what is the obvious choice...
morphism
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#8
Feb16-07, 02:37 AM
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A will commute with itself and all its powers. Also, A will commute with all multiples of Id.
dextercioby
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#9
Feb16-07, 03:04 AM
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Quote Quote by yooyo View Post
it is given A find B such that AB=BA..
what is the obvious choice...
B=A, or B=I.
robphy
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#10
Feb17-07, 06:33 AM
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special polynomials in kA, where k is a constant
matt grime
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#11
Feb17-07, 06:51 AM
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Uh? *any* polynomial in A commutes with A (dunno what the k has to do with anything).
robphy
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#12
Feb17-07, 07:05 AM
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Quote Quote by matt grime View Post
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.
AlphaNumeric
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#13
Feb17-07, 09:26 AM
P: 290
Any B sharing all of A's eigenvectors would work too wouldn't it? Or is that pretty much amounting to a polynomial of A anyway?

Or is there some subtle reason I'm missing why that wouldn't work?
matt grime
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#14
Feb17-07, 12:18 PM
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It certainly doesn't work. Not unless there is a basis of eigenvectors. It is easy to find matrices A and B with precisely one eigenvector each that contradict your hypothesis.
shadowsword
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#15
Feb19-07, 03:53 PM
P: 1
B=A^-1

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.
matt grime
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#16
Feb19-07, 04:00 PM
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Where does it say A is invertible? I don't understand your 'upper right and lower left' condition at all. 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.)
hotcommodity
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#17
Feb21-07, 04:21 PM
P: 440
So I'm assuming you want a matrix B such that a known matrix A when multiplied by B can commute, that is AB=BA. The idea is to set up a system of linear equations with all of your unknowns. First we have to specify the unknowns. If B is a 3X3 matrix then we will have a matrix containing a,b,c,d,e,f,g,h,i where these letters are the unknowns representitive of the coefficients in the B matrix. Next you want to multiply A times B, and B times A, which should give you 18 different equations. We want to treat a,b,c, etc. as if they were x1, x2, x3, etc. Remember AB=BA, which means AB - BA = 0. Now you can set up and solve for a linear system using elementary row operations. Once the linear system is in reduced row echelon form, you will see the conditions for AB=BA. I hope that helps.
matt grime
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#18
Feb21-07, 05:33 PM
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Does nobody read the f**king posts before them and try to figure out if they've answered the question or not?


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