How to show that commutative matrices form a group?

  • I
  • Thread starter Robin04
  • Start date
  • #1
Robin04
260
16
Let's say we have a given matrix ##G##. I want to find a set of ##M## matrices so that ##MG = GM## and prove that this is a group. How can I approach this problem?
 

Answers and Replies

  • #2
phyzguy
Science Advisor
5,054
2,054
Write down the definition of a group and see if you show that your matrices adhere to the definition.
 
  • Like
Likes FactChecker
  • #3
fresh_42
Mentor
Insights Author
2021 Award
17,607
18,183
Let's say we have a given matrix ##G##. I want to find a set of ##M## matrices so that ##MG = GM## and prove that this is a group. How can I approach this problem?
By more conditions, e.g. ##M=0## is possible in your setup. Do you look for an addition group?
 
  • #4
Robin04
260
16
Actually I'm not really sure how this group would look like. The set would be the ##M## matrices I suppose, and the operation? Multiplication by ##G##? How can I make ##G## a special element in the group?
 
  • #5
fresh_42
Mentor
Insights Author
2021 Award
17,607
18,183
##G## already is in the group. However, you should at least have a group operation. It looks like a multiplicative group you're heading for. I'm excited to see how you will manage the inverses without further assumptions.
 
  • #6
Robin04
260
16
Can the operation be something like the commutator? It cannot be exactly the commutator because then G wouldn't be an identity element but an aggressive element instead.
 
  • #7
fresh_42
Mentor
Insights Author
2021 Award
17,607
18,183
What is an aggressive element? Where do your matrices live in? Even the word commutator isn't defined by the above.
##\{\,M\,|\,MG=GM\,\}## is already a group, say ##H##. Then ##G\in H## and ##H## is Abelian. But I have the strong feeling that you will not be satisfied.
 
  • #8
Robin04
260
16
Aggressive element is an element that makes every other element equal to itself. Like multiplying real numbers by zero. Not sure if aggressive element is the right term in English, I learned it in Hungarian.
The matrices are in ##\mathbb{R}^{n \times n}##. And as ##G## is given, I would like to express what the ##M## matrices must look like.

##\{\,M\,|\,MG=GM\,\}## is already a group, say ##H##. Then ##G\in H## and ##H## is Abelian. But I have the strong feeling that you will not be satisfied.
I'm not sure we're on the same page. 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.

The picture I have in mind is that a group is an algebraic structure which means that there is a set and a binary operation with the following properties
- closure for that operation
- associativity
- there is an inverse element for every element in the set
- there is an identity element in the set

So far, my problem is that I don't really see how the statement "matrices commuting with a given matrix form a group" fits into this picture. What is the set, what is the operation, etc.
 
  • #9
fresh_42
Mentor
Insights Author
2021 Award
17,607
18,183
Aggressive element is an element that makes every other element equal to itself. Like multiplying real numbers by zero. Not sure if aggressive element is the right term in English, I learned it in Hungarian.
Tell me. My dictionary gave me aggresziv or eröszakos.
The matrices are in ##\mathbb{R}^{n \times n}##. And as ##G## is given, I would like to express what the ##M## matrices must look like.


I'm not sure we're on the same page. 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.
Easy: Let ##H:=\{\,M\in \mathbb{R}^{n \times n}\,|\,MG=GM\,\}##.
The picture I have in mind is that a group is an algebraic structure which means that there is a set and a binary operation with the following properties
- closure for that operation
##M,N \in H \Longrightarrow M+N \in H##
- associativity
##M+(N+P)=(M+N)+P##
- there is an inverse element for every element in the set
##M \in H \Longrightarrow -M \in H##
- there is an identity element in the set
##0\in H##
So far, my problem is that I don't really see how the statement "matrices commuting with a given matrix form a group" fits into this picture. What is the set, what is the operation, etc.
This is a rather boring solution, but if we must not invert the matrices, the more interesting multiplicative case isn't possible. The entire question including "aggressive" reminds me of algebras which are used in biology.
 
Last edited:
  • #10
Robin04
260
16
Tell me. My dictionary gave me aggresziv or eröszakos.
It's aggresszív in Hungarian. I didn't find anything for agressive element, how do you call this in English?

Easy: Let ##H:=\{\,M\in \mathbb{R}^{n \times n}\,|\,MG=GM\,\}##.

##M,N \in H \Longrightarrow M+N \in H##
##M+(N+P)=(M+N)+P##
##M \in H \Longrightarrow -M \in H##
##0\in H##

But how is this a proof that these rules also apply to commuting matrices too? Maybe it should be trivial but I don't see how the sum of two matrices that commute with ##G## also commutes with ##G##. Same for the other conditions.
 
  • #12
Stephen Tashi
Science Advisor
7,776
1,537
Let's say we have a given matrix ##G##. I want to find a set of ##M## matrices so that ##MG = GM## and prove that this is a group. How can I approach this problem?

If you interpret "is a group" to mean "is a group under the operation of multiplication" then the problem asks for a proof of a false statement. Begin by fixing the problem.

Let ##G## be an nxn matrix. Let ##S## be the set of nxn matrices such that ##m \in S## iff ##mG = Gm##.

If ##G## is the zero nxn matrix then ##S## is the set of all nxn matrices, which is not a group under the operation of multiplication.

Suppse ##G## is not an invertible matrix. The identity matrix ##I## is an element of ##S##. If ##S## were a group then it would contain ##GI= IG= G##. So ##S## would contain an matrix ##G## with no multiplicative inverse. Hence ##S## is not a multiplicative group.

----
If you interpret "is a group" to mean "is a group under the operation of addition", then, besides the trivial matters, you must show that if ##A \in S## and ##B \in S## then ##A+B \in S##.

Maybe it should be trivial but I don't see how the sum of two matrices that commute with G" role="presentation">G also commutes with G" role="presentation">G

##G(A+B) = ? = ? = (A+B)G##
 
  • #13
steenis
312
18
Let's say we have a given matrix ##G##. I want to find a set of matrices ##M## so that ##MG = GM## and prove that this is a group. How can I approach this problem?

Let ##G## be a ##n \times n## matrix

Define ##S## to be the set of ##n \times n## matrices with nonzero determinant: ##S = \{ n \times n \text{ matrix } A | \text{ } det(A) \neq 0 \}##

Define ##T## to be the set of ##n \times n## matrices with nonzero determinant that commute with ##G##: ##T = \{M \in S | \text{ } MG=GM \}##

It is easy to show that ##T## is a group

Questions:
Is it necessary that ##G \in S## ?. That is, is it necessary that ##det(G) \neq 0## ?

Does ##G## belong to ##T##, i.e., ##G \in T## ?
 
  • #14
WWGD
Science Advisor
Gold Member
6,207
7,746
If your matrix G is the identity, it will commute with non-invertible matrices, and these will not be invertible. I saw a related name commutant?
 
  • #15
WWGD
Science Advisor
Gold Member
6,207
7,746
Let ##G## be a ##n \times n## matrix

Define ##S## to be the set of ##n \times n## matrices with nonzero determinant: ##S = \{ n \times n \text{ matrix } A | \text{ } det(A) \neq 0 \}##

Define ##T## to be the set of ##n \times n## matrices with nonzero determinant that commute with ##G##: ##T = \{M \in S | \text{ } MG=GM \}##

It is easy to show that ##T## is a group

Questions:
Is it necessary that ##G \in S## ?. That is, is it necessary that ##det(G) \neq 0## ?

Does ##G## belong to ##T##, i.e., ##G \in T## ?
Doesn't every element in a group have an inverse? If Det(G)=0 , then G is not invertible.
 
  • #16
fresh_42
Mentor
Insights Author
2021 Award
17,607
18,183
If your matrix G is the identity, it will commute with non-invertible matrices, and these will not be invertible. I saw a related name commutant?
Commutator, but even this depends on the structure: ##[G,M]=GM-MG## or ##[G,M]=GMG^{-1}M^{-1}.##
 
  • #17
fresh_42
Mentor
Insights Author
2021 Award
17,607
18,183
Doesn't every element in a group have an inverse? If Det(G)=0 , then G is not invertible.
If ##\operatorname{det}G =0## then ##-G## is still the inverse :biggrin:
 
  • #19
WWGD
Science Advisor
Gold Member
6,207
7,746
If ##\operatorname{det}G =0## then ##-G## is still the inverse :biggrin:
But I thought this was a multiplicative group. An additive group has every element invertible, doesn't it, by -G itself, right?
 
  • #20
fresh_42
Mentor
Insights Author
2021 Award
17,607
18,183
Commutator is the construction, ##[,]=0## resp. ##[,]=1## the centralizer. I've never heard of commutant for centralizer, but anyway, it's a slight difference, so it depends on what is meant.
 
  • #21
steenis
312
18
Let ##G## be a ##n \times n## matrix

Define ##S## to be the set of ##n \times n## matrices with nonzero determinant: ##S = \{ n \times n \text{ matrix } A | \text{ } det(A) \neq 0 \}##

Define ##T## to be the set of ##n \times n## matrices with nonzero determinant that commute with ##G##: ##T = \{M \in S | \text{ } MG=GM \}##

It is easy to show that ##T## is a group

Questions:
Is it necessary that ##G \in S## ?. That is, is it necessary that ##det(G) \neq 0## ?

Does ##G## belong to ##T##, i.e., ##G \in T## ?

Obviously, ##T## is a multiplicative group, consisting of ##n \times n## matrices with nonzero determinant.
 
  • #22
fresh_42
Mentor
Insights Author
2021 Award
17,607
18,183
But I thought this was a multiplicative group. An additive group has every element invertible, doesn't it, by -G itself, right?
That was one of the difficulties with the OP. It hasn't been specified, and to automatically assume invertible matrices if he talks about ##\mathbb{R}^{n\times n}## is a bit of a stretch.
 
  • #23
fresh_42
Mentor
Insights Author
2021 Award
17,607
18,183
Obviously, ##T## is a multiplicative group, consisting of ##n \times n## matrices with nonzero determinant.
Obviously, this is an unsupported assumption.
 
  • #24
WWGD
Science Advisor
Gold Member
6,207
7,746
That was one of the difficulties with the OP. It hasn't been specified, and to automatically assume invertible matrices if he talks about ##\mathbb{R}^{n\times n}## is a bit of a stretch.
But if you have an aditive group then, at least in my understanding, everything commutes.
 
  • #25
fresh_42
Mentor
Insights Author
2021 Award
17,607
18,183
But if you have an aditive group then, at least in my understanding, everything commutes.
No one said it won't be a boring solution. At least it is one which doesn't depend on assumptions made by others.
 
  • #26
WWGD
Science Advisor
Gold Member
6,207
7,746
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.
 
  • #27
fresh_42
Mentor
Insights Author
2021 Award
17,607
18,183
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.)
 
  • #28
FactChecker
Science Advisor
Gold Member
7,463
3,228
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.
 
  • #29
fresh_42
Mentor
Insights Author
2021 Award
17,607
18,183
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.
 
  • #30
FactChecker
Science Advisor
Gold Member
7,463
3,228
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.
 
  • #31
Robin04
260
16
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.
 
  • #32
fresh_42
Mentor
Insights Author
2021 Award
17,607
18,183
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\,?##
 
  • #33
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
23,277
14,786
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.
 
  • #34
Robin04
260
16
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?
 
Last edited:
  • #35
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
23,277
14,786
##(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?
 

Suggested for: How to show that commutative matrices form a group?

Replies
27
Views
2K
  • Last Post
Replies
0
Views
359
Replies
3
Views
287
Replies
1
Views
547
MHB Matrices
  • Last Post
Replies
1
Views
309
  • Last Post
Replies
5
Views
682
Replies
14
Views
1K
Replies
30
Views
2K
  • Last Post
Replies
13
Views
1K
Top