Matrices that commute with A

[James Crook]

We have the 2x2 matrix A:

6 7

4 5 

We know that it commutes with I, with A, with Inv(A), where Inv(A) is the inverse:

-2  3/2
1  -1/2

(Kelsey)

And A commutes with any constant multiples of these.  We also know that A commutes with A^2 (i.e A squared ) because

A(AA) = (AA)A

...because matrix multiplication is associative.  And

A(AAA) = (AAA)A

so it's clear all powers of A commute with A.  In fact if A commutes with B then A commutes with all powers of B.  We can just move the B's from one side to the other one at a time.  Here we're easily showing that A commutes with B cubed:

ABBB = BABB = BBAB = BBBA

So since we know A commutes with its inverse, A commutes with powers of its inverse too.

We can regard the identity matrix I as A to the power of zero.  We can regard Inv(A) as A to the power of -1 and we can regard Inv(A) squared as A to the power of -2.

So so far we've seen that A commutes with powers of itself, and also all constant multiples of those.

Are we done?  Have we found all matrices that commute with A?

--James.

[James Crook]

Are there any other matrices that commute with A?

--James.

[Shri R Ganeshram]

with all powers of A and all powers of Inv(A)
________________________________________
From: l2lea...@googlegroups.com [l2lea...@googlegroups.com] On Behalf Of James Crook [james....@gmail.com]
Sent: Wednesday, June 22, 2011 2:33 PM
To: l2lea...@googlegroups.com
Subject: Re: Matrices that commute with A

Are there any other matrices that commute with A?

--James.

[James Crook]

We already know that A commutes with all powers of A and all powers of Inv(A), and the identity matrix 'I' (which we count as A to the zero) and all multiples of these by a constant....

So have we accounted for all matrices that commute with A or are there any more possibilities?  That's what I am asking.

--James.

[Shri R Ganeshram]

a b
c d
6a+4b=6a+7c
7a+5b=6a+7b
6c+4d=4a+5c
7c+5d=4b+5d

We find all solutions from solving the above system. The system was obtained by taking a matrix B, a b and multiplying it on both sides by A, and setting each part of the resulting matrices equal.
c d
We can obtain all solutions by taking the following matrix to reduced row echelon form, after setting each equation to zero, first column is a, second b, etc:

0 4 -7 0 0
1 -2 0 0 0
-4 0 1 4 0
0 -4 7 0 0

~rref

http://www3.wolframalpha.com/Calculate/MSP/MSP94519g3iih56a1g9960000023i94hcf679f30hc?MSPStoreType=image/gif&s=38&w=225&h=76
http://www3.wolframalpha.com/Calculate/MSP/MSP94819g3iih56a1g9960000058bgc53i5ha3df3e?MSPStoreType=image/gif&s=38&w=161&h=76

^ see links for matrices
we have a free variable, since we only have 3 rows whilst we have 4 variables. Since "d" is common in all 3 equations, we say "d" is our free variable and represent "a", "b", and "c" in terms of "d".
a=14/13d
b=7/13d
c=4/13d

so any matrix of the form

d/13*(14 7)
(4 13) will work.

(I think)

________________________________________
From: l2lea...@googlegroups.com [l2lea...@googlegroups.com] On Behalf Of James Crook [james....@gmail.com]

Sent: Wednesday, June 22, 2011 4:10 PM

To: l2lea...@googlegroups.com
Subject: Re: Matrices that commute with A

We already know that A commutes with all powers of A and all powers of Inv(A), and the identity matrix 'I' (which we count as A to the zero) and all multiples of these by a constant....

So have we accounted for all matrices that commute with A or are there any more possibilities? That's what I am asking.

--James.

On 22 June 2011 20:41, Shri R Ganeshram <sh...@mit.edu<mailto:sh...@mit.edu>> wrote:
with all powers of A and all powers of Inv(A)
________________________________________

From: l2lea...@googlegroups.com<mailto:l2lea...@googlegroups.com> [l2lea...@googlegroups.com<mailto:l2lea...@googlegroups.com>] On Behalf Of James Crook [james....@gmail.com<mailto:james....@gmail.com>]

Sent: Wednesday, June 22, 2011 2:33 PM

To: l2lea...@googlegroups.com<mailto:l2lea...@googlegroups.com>

Subject: Re: Matrices that commute with A

Are there any other matrices that commute with A?

--James.

[James Crook]

Anyone like to comment on Shri's solution?

Does it cover the examples that we already know work, such as I and A itself?

--James.

[James Crook]

Abe,

Shri made a small slip in writing down the equations that a,b,c,d must satisfy and that is why the answer he got at the end is not a complete answer.  It's why it doesn't include the identity, and we do know that the identity should be a possible solution.  That should have been an alarm bell that the solution was not right.  There is also an earlier alarm bell, in that the initial equations involve 'a' five times and 'b' just three.  Because of the symmetry of the initial set up we'd expect to see each variable 4 times.

Abe, please would you write the correct initial equations?  I'll then take it from there.  Thanks.

[Shri Ganeshram]

The second equation should have 6b+7d... sorry guys

[James Crook]

So:

6a+4b=6a+7c
7a+5b=6b+7d
6c+4d=4a+5c
7c+5d=4b+5d

Cancelling a few...

     4b=7c
     7a=b+7d
 c+4d=4a
     7c=4b

Anyone like to finish this off, remembering that every time we get two equations the same we can cross one out, and know we have a free choice in one of the variables.

--James. 

[James Crook]

4 days to go on this problem.  Any takers for completing it?

[James Crook]

3 days to go on this problem.

[James Crook]

2 days to go.

If time is up and no one has done it then I will do it - but not having it done already is slowing us down.  Please, someone, help us move on a little faster than this...

--James.

[Jonathan Joo]

Sorry for the multitude of emails-

I tried solving for the variables, but the four equations all canceled out- does that mean that all numbers are solutions? Or was I doing something wrong?

-Jonathan

---

From: James Crook <james....@gmail.com>
To: l2lea...@googlegroups.com
Sent: Monday, June 27, 2011 4:26 PM

Subject: Re: Matrices that commute with A

[James Crook]

There is a lot of cancelling, but they don't all cancel out completely.  You should be able to find what line you were over-zealous in cancelling.  For example, we have 4b=7c and 7c=4b, they are the same equations, so we can get rid of one, that's OK to do.  Given the answer you got, your most likely slip is that you got rid of both.  So you then said that b and c can be anything.  But in fact even at this stage we know for sure that b = (7c)/4 if the matrices commute. 

[In particular this already shows that the matrix

a, 1
1, d

will not commute with A, no matter what choices we make for a and d]


If on checking the steps this is still not working for you, write out the steps you got in an e-mail so that I can diagnose.  I'd like to also see you have another go, but if you'd rather not do that, show us the steps you took, and nominate someone else in the group to take over the problem, to fix the steps and complete it.

Jonathan - I am really glad you have had a go at this problem and have the courage to report what you got.  For this study group that is 100% better than sitting on the sidelines waiting for someone else.  Thanks for reporting back.

--James.

[Jonathan Joo]

7c=4b

 7a=b+7d-----b=7a-7d
 c+4d=4a-----c=4a-4d

7(4a-4d)=4(7a-7d)

28a-28d=28a-28d

0=0

I keep on getting equations like this that end up being the same on both sides... I'm probably missing something easy though. Anyone else want to take over?

-Jonathan

---

From: James Crook <james....@gmail.com>
To: l2lea...@googlegroups.com

Sent: Wednesday, June 29, 2011 5:32 AM

[James Crook]

b=7a-7d

c=4a-4d

So far so good. Jonathan - you then went on to prove that 0=0, which of course is correct, but embarrassingly doesn't get us any closer to the answer.  Writing out one extra step helps us get through to an answer: 

4b=28a-28d

7c=28a-28d

These are the same equation as each other (because 4b=7c) so we can throw one of them out.  We don't throw them both out.  So now we are down from 3 equations in 4 unknowns to 2 equations in 4 unknowns.

7c=4b

7c=28a-28d

We can make 'c' and 'a' arbitrary and deduce 'b' and 'd', we can choose 'a' and 'd' and deduce 'b' and 'c', we can choose 'b' and 'd' and deduce 'a' and 'c'...  we've therefore plenty of choices for how to present the answer.  Please, someone, make a choice and give the answer....

--James.