I How to show that commutative matrices form a group?

  • Thread starter Robin04
  • Start date
247
14
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?
 

phyzguy

Science Advisor
4,295
1,281
Write down the definition of a group and see if you show that your matrices adhere to the definition.
 

fresh_42

Mentor
Insights Author
2018 Award
11,615
8,094
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?
 
247
14
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?
 

fresh_42

Mentor
Insights Author
2018 Award
11,615
8,094
##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.
 
247
14
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.
 

fresh_42

Mentor
Insights Author
2018 Award
11,615
8,094
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.
 
247
14
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 learnt 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.
 

fresh_42

Mentor
Insights Author
2018 Award
11,615
8,094
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 learnt 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:
247
14
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.
 

Stephen Tashi

Science Advisor
6,765
1,102
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##
 
46
17
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## ?
 

WWGD

Science Advisor
Gold Member
4,570
1,979
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?
 

WWGD

Science Advisor
Gold Member
4,570
1,979
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.
 

fresh_42

Mentor
Insights Author
2018 Award
11,615
8,094
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}.##
 

fresh_42

Mentor
Insights Author
2018 Award
11,615
8,094
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:
 

WWGD

Science Advisor
Gold Member
4,570
1,979
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?
 

fresh_42

Mentor
Insights Author
2018 Award
11,615
8,094
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.
 
46
17
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.
 

fresh_42

Mentor
Insights Author
2018 Award
11,615
8,094
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.
 

fresh_42

Mentor
Insights Author
2018 Award
11,615
8,094
Obviously, ##T## is a multiplicative group, consisting of ##n \times n## matrices with nonzero determinant.
Obviously, this is an unsupported assumption.
 

WWGD

Science Advisor
Gold Member
4,570
1,979
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.
 

fresh_42

Mentor
Insights Author
2018 Award
11,615
8,094
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.
 

Want to reply to this thread?

"How to show that commutative matrices form a group?" You must log in or register to reply here.

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

Replies
3
Views
1K
Replies
1
Views
2K
Replies
5
Views
3K
Replies
1
Views
417
  • Posted
Replies
7
Views
4K
Replies
5
Views
10K
Replies
7
Views
911
  • Posted
Replies
4
Views
2K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top