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Robin04

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Robin04

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- #2

phyzguy

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By more conditions, e.g. ##M=0## is possible in your setup. Do you look for an addition group?

- #4

Robin04

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- #6

Robin04

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##\{\,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

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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.##\{\,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.

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

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Tell me. My dictionary gave me aggresziv or eröszakos.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.

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

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

##M \in H \Longrightarrow -M \in H##- there is an inverse element for every element in the set

##0\in H##- there is an identity element in the set

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.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.

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- #10

Robin04

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It's aggresszív in Hungarian. I didn't find anything for agressive element, how do you call this in English?Tell me. My dictionary gave me aggresziv or eröszakos.

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.

- #11

Stephen Tashi

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I think it's an "absorbing element". https://en.wikipedia.org/wiki/Absorbing_elementIt's aggresszív in Hungarian. I didn't find anything for agressive element, how do you call this in English?

- #12

Stephen Tashi

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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

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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

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- #15

WWGD

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Doesn't every element in a group have an inverse? If Det(G)=0 , then G is not invertible.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## ?

- #16

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

- #17

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If ##\operatorname{det}G =0## then ##-G## is still the inverseDoesn't every element in a group have an inverse? If Det(G)=0 , then G is not invertible.

- #18

WWGD

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

- #19

WWGD

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

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- #21

steenis

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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

- #22

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

- #23

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Obviously, this is an unsupported assumption.Obviously, ##T## is a multiplicative group, consisting of ##n \times n## matrices withnonzerodeterminant.

- #24

WWGD

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But if you have an aditive group then, at least in my understanding, everything commutes.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.

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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.But if you have an aditive group then, at least in my understanding, everything commutes.

- #26

WWGD

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

- #27

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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.)Not likely. Hungarians are top Math people. He would probably blacklist him if he asked a simple question.

- #28

FactChecker

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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

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He already said it:

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.

Since we cannot talk to his professor, we can only take what we have, and there is a solution with a "yes" as answer.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.

- #30

FactChecker

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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.He already said it:

- #31

Robin04

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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.

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- #33

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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.

- #34

Robin04

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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.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.

##\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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- #35

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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?

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