The webpage title could be Matrix Equivalence Theorem: Proving A=γI from AB=BA

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We consider a matrix A, NxN.Show that if for every NxN matrix B we have AB=BA,then the matrix A is of the form:A=γI.
When I first looked at this exercise,the first I thought was to assume that A=γI is true and replace A in AB=BA=>γB=Bγ=γB,which is true.But then I noticed that I have "=>" and not "<=>",so,since it just can't be that easy and simple,I guessed that's wrong.Any thoughts?
 
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chester20080 said:
We consider a matrix A, NxN.Show that if for every NxN matrix B we have AB=BA,then the matrix A is of the form:A=γI.
When I first looked at this exercise,the first I thought was to assume that A=γI is true and replace A in AB=BA=>γB=Bγ=γB,which is true.But then I noticed that I have "=>" and not "<=>",so,since it just can't be that easy and simple,I guessed that's wrong.Any thoughts?

Say the problem is stated If ##P \implies Q##.

What you did was the converse: ##Q \implies P##, which isn't equivalent.

The contrapositive IS equivalent however: ~##Q \implies## ~##P##. (Latex was messing up on me, sorry if you saw this before my edit)
 
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scurty, if you want to put a twiddle inside of latex you can use \sim
 
chester20080 said:
When I first looked at this exercise,the first I thought was to assume that A=γI is true and replace A in AB=BA=>γB=Bγ=γB,which is true.
That's trivial. The problem at hand is much harder. You need to show that the only way AB=BA is true for all NxN matrices B is if A is some scalar times the identity matrix.
 
Office_Shredder said:
scurty, if you want to put a twiddle inside of latex you can use \sim
Or \neg. I edited scurty's post to use \neg just before you edited it to use \sim. Personally I like \neg better because the command says what you want, while \sim seems to be exactly the other way around.
 
Yep,I figured out that was incorrect...So any tip on how to proceed?
 
When they say a property is true for ALL matrices B, the best way to approach the problem is to think of matrices B that give you a lot of information about what A is. If AB = BA and B is a specific matrix, you get a lot of information about how the entries of A are related. Ideally you would pick a matrix B which makes the information very simple, like specific entries being equal to zero, or two different entries being equal. Try playing around with some matrices B that aren't too hard to do calculations with.
 
Thanks Office_Shredder and DH. I used Detexify but nothing was showing up for me and \~ wasn't working. Now I know for the future!
 
I don't know how to proceed.I tried some B matrices to see what is going on with A.Then I assumed A=γI,so all fields are zero except for those on the main diagonal.What I got was that the elements of the diagonal were all the same,as expected.But I don't know how to generalize that and how the result to come out in a physical way.Should I consider random A,B matrices and assume that A=γI and to make zero all the non-main-diagonal elements of A in order to have all the main diagonal elements the same?But isn't that exactly the same with "=>"?Any further help?
 
  • #10
You are looking at the problem the wrong way. Don't start with A=γI. Start with the given condition, that AB=BA for all matrices B. You have to prove that A=γI is the only solution that works for all matrices B.

Hint: Look at the simplest of non-zero matrices B. For example, in the case of 2x2 matrices, look at the matrix ##B=\begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}##. What does this tell you about A? You need one more such simple matrix B to tell you that A must be a scalar times the 2x2 identity matrix.
 
  • #11
But the exercise says to show that A is of that form,not that A has ONLY that form.Why should I care if A is of any other form?And from 2x2 how can I generalize it to NxN?
 
  • #12
chester20080 said:
But the exercise says to show that A is of that form,not that A has ONLY that form.Why should I care if A is of any other form?And from 2x2 how can I generalize it to NxN?

You still have it exactly backwards. You start with some NxN matrix about which almost nothing is know---no form, or anything is given to you. You are, however, told one bit of information, namely that AB = BA for all NxN matrices B. From that you are required to hammer down A to a special form, namely, a scalar times the identity matrix. Thefore, A = yI is not input, it is output.

And, to answer your other question: you generalize it by sitting down and doing it! If you don't see how, then try the 3x3 case first. Don't agonize over it; just get started.
 
  • #13
chester20080 said:
But the exercise says to show that A is of that form,not that A has ONLY that form.Why should I care if A is of any other form?And from 2x2 how can I generalize it to NxN?

You are misreading the problem statement. It does not say to show that if the N×N matrix A is of the form A=γI then AB=BA for all N×N matrices B. It instead says to show that if AB=BA for all N×N matrices B then A is of the form A=γI. Another way to read "is of the form" is "must be of the form". Showing that AB=BA if A is a scalar times the identity does not show that A must be of that form because if P then Q is not the same as if Q then P.
 
  • #14
Office_Shredder said:
scurty, if you want to put a twiddle inside of latex you can use \sim
A bit off-topic, but I'm curious about a couple of things.

"twiddle" = "tilde"?
\neg I get, but what is \sim short for?
 
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