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I How to show that commutative matrices form a group?

  1. Oct 10, 2018 #1
    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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  3. Oct 10, 2018 #2

    phyzguy

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    Write down the definition of a group and see if you show that your matrices adhere to the definition.
     
  4. Oct 10, 2018 #3

    fresh_42

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    By more conditions, e.g. ##M=0## is possible in your setup. Do you look for an addition group?
     
  5. Oct 10, 2018 #4
    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?
     
  6. Oct 10, 2018 #5

    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.
     
  7. Oct 10, 2018 #6
    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.
     
  8. Oct 10, 2018 #7

    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.
     
  9. Oct 10, 2018 #8
    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.

    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.
     
  10. Oct 10, 2018 #9

    fresh_42

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    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##
    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: Oct 10, 2018
  11. Oct 11, 2018 #10
    It's aggresszív in Hungarian. I didn't find anything for agressive element, how do you call this in English?

    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. Oct 11, 2018 #11

    Stephen Tashi

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  13. Oct 11, 2018 #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##.

    ##G(A+B) = ? = ? = (A+B)G##
     
  14. Oct 11, 2018 #13
    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## ?
     
  15. Oct 11, 2018 #14

    WWGD

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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?
     
  16. Oct 11, 2018 #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.
     
  17. Oct 11, 2018 #16

    fresh_42

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    Commutator, but even this depends on the structure: ##[G,M]=GM-MG## or ##[G,M]=GMG^{-1}M^{-1}.##
     
  18. Oct 11, 2018 #17

    fresh_42

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    If ##\operatorname{det}G =0## then ##-G## is still the inverse :biggrin:
     
  19. Oct 11, 2018 #18

    WWGD

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  20. Oct 11, 2018 #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?
     
  21. Oct 11, 2018 #20

    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.
     
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