How to make a matrix B such that AB=BA

[yooyo]

B is not the identity matrix or the zero matrix
suppose both A and B are 3X3
I don't have clue on this..any hits?

 

[d_leet]

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]

how abouit 2Id?

 

[yooyo]

what do you mean by 2Id? 2xI?

 

[matt grime]

There is an obvious choice for B (well infinitely many) that is not Id or 0. What will A always commute with?

 

[HallsofIvy]

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

 

[yooyo]

it is given A find B such that AB=BA..
what is the obvious choice...

 

[morphism]

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

 

[dextercioby]

it is given A find B such that AB=BA..
what is the obvious choice...

B=A, or B=I.

 

[robphy]

special polynomials in kA, where k is a constant

 

[matt grime]

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

 

[robphy]

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]

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]

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]

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]

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]

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]

Does nobody read the f**king posts before them and try to figure out if they've answered the question or not?

 

[hotcommodity]

Does nobody read the f**king posts before them and try to figure out if they've answered the question or not?

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."

 

[matt grime]

Several suggestions have been made that lead to infinitely many B, none of which require the setting up or solution of anything as tedious as simultaneous equations.

 

[D H]

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 ...

 

[matt grime]

I would also like to point out that hotcommodity's suggested solution is precisely the first one given in the thread in post 2 by d_leet as well, which is what really annoyed me even more than ignoring the better and easier methods.

 

[hotcommodity]

I would also like to point out that hotcommodity's suggested solution is precisely the first one given in the thread in post 2 by d_leet as well, which is what really annoyed me even more than ignoring the better and easier methods.

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.

 

[D H]

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 B=\sum_n a_nA^n.

Edit: There are nine unknowns, not six.

 

[hotcommodity]

I made up a matrix A to show my example, the word document shows the example with maple. Sorry I'm not familiar with the equation in your last post, I'm still learning myself.

 

[matt grime]

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...

 

[hotcommodity]

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

 

[HallsofIvy]

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

 

[hotcommodity]

Oh I see, lol.

 

[mathwonk]

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.

 

[leon1127]

B = 0
B = I
B = A
B = A^-1

There are more I suppose

 

[DeadWolfe]

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?

 

[navidnfk]

I have a Idea How to Find B that AB=BA!

 

[HallsofIvy]

Without knowing A?

 

[trambolin]

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.

 

[hotcommodity]

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

 

[JasonRox]

I stand by matt_grime. It's frustrating although I'm probably that person from time to time.

matt_wonk, that's a boring question. Um...