# Theoretical matrix problem

1. Dec 18, 2013

### chester20080

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?

Last edited: Dec 18, 2013
2. Dec 18, 2013

### scurty

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)

Last edited: Dec 18, 2013
3. Dec 18, 2013

### Office_Shredder

Staff Emeritus
scurty, if you want to put a twiddle inside of latex you can use \sim

4. Dec 18, 2013

### D H

Staff Emeritus
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.

5. Dec 18, 2013

### D H

Staff Emeritus
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.

6. Dec 18, 2013

### chester20080

Yep,I figured out that was incorrect...So any tip on how to proceed?

7. Dec 18, 2013

### Office_Shredder

Staff Emeritus
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.

8. Dec 18, 2013

### scurty

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!

9. Dec 19, 2013

### chester20080

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. Dec 19, 2013

### D H

Staff Emeritus
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. Dec 19, 2013

### chester20080

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. Dec 19, 2013

### Ray Vickson

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. Dec 19, 2013

### D H

Staff Emeritus
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. Dec 19, 2013

### Staff: Mentor

A bit off-topic, but I'm curious about a couple of things.

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