How to show that commutative matrices form a group?

[Robin04]

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]

Write down the definition of a group and see if you show that your matrices adhere to the definition.

 

[fresh_42]

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?

 

[Robin04]

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]

$G$ 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.

 

[Robin04]

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]

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.

 

[Robin04]

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.

 

[fresh_42]

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.

 

[Robin04]

Tell me. My dictionary gave me aggresziv or eröszakos.

It's aggresszív in Hungarian. I didn't find anything for aggressive 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]

It's aggresszív in Hungarian. I didn't find anything for aggressive element, how do you call this in English?

I think it's an "absorbing element". https://en.wikipedia.org/wiki/Absorbing_element

 

[Stephen Tashi]

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

$G(A+B) = ? = ? = (A+B)G$

 

[steenis]

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]

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]

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]

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]

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

 

[WWGD]

Commutator, but even this depends on the structure: $[G,M]=GM-MG$ or $[G,M]=GMG^{-1}M^{-1}.$

Is this the same as Centralizer?https://en.wikipedia.org/wiki/Centralizer_and_normalizer (See the first few lines, where it is also nammed commutant)

 

[WWGD]

If $\operatorname{det}G =0$ then $-G$ is still the inverse

But I thought this was a multiplicative group. An additive group has every element invertible, doesn't it, by -G itself, right?

 

[fresh_42]

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.

 

[steenis]

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]

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]

Obviously, $T$ is a multiplicative group, consisting of $n \times n$ matrices with nonzero determinant.

Obviously, this is an unsupported assumption.

 

[WWGD]

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]

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.

 

[WWGD]

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]

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]

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]

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]

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.