I How to show that commutative matrices form a group?

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WWGD

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No one said it won't be a boring solution. At least it is one which doesn't depend on assumptions made by others.
Not likely. Hungarians are top Math people. He would probably blacklist him if he asked a simple question.
 

fresh_42

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Not likely. Hungarians are top Math people. He would probably blacklist him if he asked a simple question.
As he spoke about absorbing elements, I couldn't but think of the world of algebras, since in groups, there are no such elements. But as soon as we talk about algebras, we are in the additive world. That doesn't imply, that ##\{\,M\,|\,[G,M]=0\,\}## hasn't an additional multiplicative structure, however, not a multiplicative group, an additive one (subspace) with additional multiplicative properties (e.g. ideal). So all in all, the additive component doesn't make it trivial, rather does it fit better to the rest of what has been given, than automatically make hidden assumptions. (Sorry, I'm really no friend of reading more out of a post than what is actually given. The question about the operation was the first we have asked, which was basically answered by: any.)
 

FactChecker

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Is this a class homework problem? It sounds like one. If so, you need to show some work and follow the template.

In any case, rather than getting caught up in a lot of discussions, you should go one-by-one through the required properties of a group and see if you can prove them or not. If you have problems, then you should ask more specific questions.
 

fresh_42

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Is this a class homework problem? It sounds like one. If so, you need to show some work and follow the template.

In any case, rather than getting caught up in a lot of discussions, you should go one-by-one through the required properties of a group and see if you can prove them or not. If you have problems, then you should ask more specific questions.
He already said it:
I've been learning linear algebra for one month only, so I might not see your point immidiately. My teacher mentioned this problem and I found it interesting.
Since we cannot talk to his professor, we can only take what we have, and there is a solution with a "yes" as answer.
 

FactChecker

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He already said it:
Yes, I assumed so but did not feel like looking to see if he eventually said it. But that should not excuse him from not not following the template and showing his work. The fact remains that he should go one-by-one through the required properties and ask a specific question if he has a problem. @phyzguy said that in post #2, but that appears to have been ignored.
 
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I agree that the problem wasn't specific enough because I didn't really understand it myself either. Now I think I'm starting to get a clearer picture about it thanks to this discussion. First I didn't understand why do I have to choose between additive and multiplicative groups but now I see why is this a question. This wasn't specified by my teacher but I have a partial answer as I worked a bit on the problem.

I started with simple 2x2 matrices. Let's say we have ## G = \begin{pmatrix} a & b \\c & d \end{pmatrix}##, where ##a,b,c,d \in \mathbb{R}##.
We're looking for all possible ## M = \begin{pmatrix} \alpha & \beta \\\gamma & \delta \end{pmatrix}##, ##\alpha,\beta,\gamma,\delta \in \mathbb{R}## such that ##GM = MG##
I did both multiplications, set up a system of equations and solved it.
##\alpha = \frac{\gamma (a-d)}{c}##
##\beta = \frac{\gamma b}{c}##
##\gamma \in \mathbb{R}##
##\delta = 0##
These ##M## matrices form a group with addition. Now what I don't know is how to generalize this to ##n \times n## matrices and I'm not sure if they form a group with other operations too. They don't with multiplication.
Also ##M## cannot be equal to ##G## but ##GG=GG## is necessarily true.
 

fresh_42

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Given any matrix ##G## and a set ##X:=\{\,M\,|\,MG=GM\,\}##. Now what is ##(M-N)\cdot G - G\cdot (M-N)## and what does this mean for the properties of ##X\,?##
 

PeroK

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I started with simple 2x2 matrices. Let's say we have ## G = \begin{pmatrix} a & b \\c & d \end{pmatrix}##, where ##a,b,c,d \in \mathbb{R}##.
We're looking for all possible ## M = \begin{pmatrix} \alpha & \beta \\\gamma & \delta \end{pmatrix}##, ##\alpha,\beta,\gamma,\delta \in \mathbb{R}## such that ##GM = MG##
I did both multiplications, set up a system of equations and solved it.
##\alpha = \frac{\gamma (a-d)}{c}##
##\beta = \frac{\gamma b}{c}##
##\gamma \in \mathbb{R}##
##\delta = 0##
These ##M## matrices form a group with addition. Now what I don't know is how to generalize this to ##n \times n## matrices and I'm not sure if they form a group with other operations too. They don't with multiplication.
Also ##M## cannot be equal to ##G## but ##GG=GG## is necessarily true.
There's something not right here. How can we need ##\delta = 0##? The identity matrix, for which ##\delta = 1##, commutes with everything. And, how can we have "##M## cannot be equal to ##G##"? By your own analysis ##G## always commutes with itself.
 
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Given any matrix ##G## and a set ##X:=\{\,M\,|\,MG=GM\,\}##. Now what is ##(M-N)\cdot G - G\cdot (M-N)## and what does this mean for the properties of ##X\,?##
##(M-N)\cdot G - G\cdot (M-N) = GN-NG##. But what is ##N##? Element of ##X## or an arbitrary matrix?

There's something not right here. How can we need ##\delta = 0##? The identity matrix, for which ##\delta = 1##, commutes with everything. And, how can we have "##M## cannot be equal to ##G##"? By your own analysis ##G## always commutes with itself.
You're right. I was too tired last night so I used a linear equation solver and it seems it has some bugs when it comes to dealing with parameters. I solved it on my own and it seems I got it right this time.

##\alpha \in \mathbb{R}##
##\beta \in \mathbb{R} ##
##\gamma = \beta \frac{c}{b}##
##\delta = \beta\frac{d-a}{b}+\alpha##
These matrices form a group too, and ##I,G## are also in it.
How to generalize this to nxn matrices?
 
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PeroK

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##(M-N)\cdot G - G\cdot (M-N) = GN-NG##. But what is ##N##? Element of ##X## or an arbitrary matrix?


You're right. I was too tired last night so I used a linear equation solver and it seems it has some bugs when it comes to dealing with parameters. I solved it on my own and it seems I got it right this time.

##\alpha \in \mathbb{R}##
##\beta \in \mathbb{R} ##
##\gamma = \beta \frac{c}{b}##
##\delta = \beta\frac{d-a}{b}+\alpha##
These matrices form a group too, and ##I,G## are also in it.
How to generalize this to nxn matrices?
I'd still question what happens if ##b = 0## in those equations.

In general, solving equations for each matrix entry is not a very efficient way to demonstrate the properties of a matrix. I think you need to start considering the properties of the matrices involved more directly.

Are you learning this on your own?
 

fresh_42

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##(M-N)\cdot G - G\cdot (M-N) = GN-NG##. But what is ##N##? Element of ##X## or an arbitrary matrix?
Both have been meant to commutate with ##G##, so ##M,N\in X##.
The point is, that this is sufficient to show what you need: ##0\in X\; , \; -N \in X\; , \;M+N\in X## and even ##G \in X##, although not of interest for the group property.
Next it shows, that all these have nothing to do with a specific dimension or the solution of your system of linear equations. All you additionally need is that matrix addition is associative.
 

Stephen Tashi

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How to generalize this to nxn matrices?
Forget about the particulars of matrices. What you want to prove holds for any mathematical system that obeys the associative law (posts #12, #32)
 
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I would also learn how to write matrix multiplication using Summation notation.
 

WWGD

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##(M-N)\cdot G - G\cdot (M-N) = GN-NG##. But what is ##N##? Element of ##X## or an arbitrary matrix?


You're right. I was too tired last night so I used a linear equation solver and it seems it has some bugs when it comes to dealing with parameters. I solved it on my own and it seems I got it right this time.

##\alpha \in \mathbb{R}##
##\beta \in \mathbb{R} ##
##\gamma = \beta \frac{c}{b}##
##\delta = \beta\frac{d-a}{b}+\alpha##
These matrices form a group too, and ##I,G## are also in it.
How to generalize this to nxn matrices?
But this too seems problematic. If ##\beta ## is in ##\mathbb R## then ##\gamma ## can be anything.
 

pasmith

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The more useful proposition is that the set [itex]X_G[/itex] of nxn matrices which commute with a specific nxn matrix [itex]G[/itex] form a ring under matrix addition and multiplication. Now this ring is a subring of the ring of all nxn matrices, so we only have to check closure of addition and multiplication and that [itex]X_G[/itex] contains identities and additive inverses. This requires proving the following:

1) If M and N commute with G then so does M + N.
2) If M and N commute with G then so does MN.
3) The nxn zero matrix commutes with G.
4) The nxn identity matrix commutes with G.
5) If M commutes with G then so does -M.

All of these can be proved directly from ring axioms and the definition of "commute". Thus (1) follows from distributivity of multiplication over addition, and (2) follows from associativity of multiplication.
 

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