I 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?

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phyzguy

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

fresh_42

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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

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$G$ 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

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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 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

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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.

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Robin04

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

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.

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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

Gold Member
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

Gold Member
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

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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

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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

Gold Member

WWGD

Gold Member
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

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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

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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

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

WWGD

Gold Member
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

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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.

"How to show that commutative matrices form a group?"

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