# Does a set of matrices form a ring? Or what is the algebraic structure?

• I
In summary,There are two ways that a set of matrices can be everything: as a set of linear representations in a vector space, or as the set of all invertible matrices.

Square matrices are closed under addition and their own form of multiplication, but in general do not commute.

What algebraic structure then describes this, along with polynomials of matrices and allows us to amend with other operations, such as differentiation or integration defined on these objects? Would the appropriate algebraic structure be a ring or a field?

Yes, the set of all ## n \times n ## Matrices over a ring R are a ring, not a field, since your general matrix is not invertible under multiplication. notice this, matrices being square, is needed to define multiplication. The invertible matrices form a group under multiplication, not a field, because they're not even closed under addition.

Last edited by a moderator:
dextercioby and topsquark
WWGD said:
Yes, the set of all ## n \times n ## Matrices over a ring R are a ring, not a field, since your general matrix is not invertible under multiplication. notice this, matrices being square, is needed to define multiplication. The invertible matrices form a group under multiplication, not a field, because they're not even closed under addition.
Really? They're not closed over addition? I can't think of an example where I don't have an additive inverse.

Really? They're not closed over addition? I can't think of an example where I don't have an additive inverse.
not the existence of an additive inverse, but closure under addition. Try Id + -Id ; Id the identity. Is it invertible, this sum of invertible matrices?

topsquark
WWGD said:
Yes, the set of all ## n \times n ## Matrices over a ring R are a ring, not a field, since your general matrix is not invertible under multiplication. notice this, matrices being square, is needed to define multiplication. The invertible matrices form a group under multiplication, not a field, because they're not even closed under addition.
Just for nitpicking: they form a field in case ##R## is one and ##n=1.##

It is not pure nitpicking. The center of the full matrix ring over a field is the diagonal matrices with the same entry (Zassenhaus, IIRC). Hence, the field plays indeed a role beside the trivial case.

This means:

A set of matrices can be all of them:
• multiplicative group
• vector space
• ring
• module
• algebra
• division algebra
• field
Whatever you like. A set of matrices can be everything. You need to be way more specific in your questions.

Last edited:
dextercioby and topsquark
The usual process is the other way around. E.g. given a group ##G##, then find and classify all group homomorphisms ##G\longrightarrow GL(V)## for a vector space ##V.## The same question arises in all categories, not only groups. E.g. for a Lie algebra ##\mathfrak{g}## we are interested in all Lie algebra homomorphisms ##\mathfrak{g}\longrightarrow \mathfrak{gl}(V).##

The linear representations (matrices) are normally better understood than the original structure. They can therefore be used to gain information about that structure.

topsquark
WWGD said:
not the existence of an additive inverse, but closure under addition. Try Id + -Id ; Id the identity. Is it invertible, this sum of invertible matrices?
Okay, thank you for clarifying.

WWGD
fresh_42 said:
The usual process is the other way around. E.g. given a group ##G##, then find and classify all group homomorphisms ##G\longrightarrow GL(V)## for a vector space ##V.## The same question arises in all categories, not only groups. E.g. for a Lie algebra ##\mathfrak{g}## we are interested in all Lie algebra homomorphisms ##\mathfrak{g}\longrightarrow \mathfrak{gl}(V).##

The linear representations (matrices) are normally better understood than the original structure. They can therefore be used to gain information about that structure.
The other way around compared to what?

topsquark
WWGD said:
The other way around compared to what?
... to the mess in post #1.

topsquark
WWGD said:
I don't understand how @topsquark comes into play here, but I think that @askmathquestions isn't really interested in learning a certain subject.

We always have to make guesses to make his questions answerable. E.g. you made the assumption that ...
Would the appropriate algebraic structure be a ring or a field?
... meant ...
WWGD said:
Yes, the set of all ## n \times n ## Matrices over a ring R are a ring, not a field ...
... in order to be able to give an answer.

PeroK and topsquark
As it happens I could use a good source for that. :)

I know at least the basics but it's always good to have a good source to refer to.

-Dan

WWGD and fresh_42
topsquark said:
As it happens I could use a good source for that. :)

I know at least the basics but it's always good to have a good source to refer to.

-Dan
Sorry :) , I mistook you for askmathquestions

Last edited:
topsquark
fresh_42 said:
I don't understand how @topsquark comes into play here, but I think that @askmathquestions isn't really interested in learning a certain subject.

We always have to make guesses to make his questions answerable. E.g. you made the assumption that ...
I'm almost always interested in learning new subjects, why are you being elitist?
I'm not a born-rich mathematician who's had unlimited time and money their whole life, I don't have the background to know how to perfectly phrase every single detail of every possible conceivable single question, sorry for not being rich like you apparently. Even if you're not rich, you've certainly internalized the pervasive elitism that has persisted in mathematics for hundreds of years.

I'm almost always interested in learning new subjects, why are you being elitist?
I'm not a born-rich mathematician who's had unlimited time and money their whole life, I don't have the background to know how to perfectly phrase every single detail of every possible conceivable single question, sorry for not being rich like you apparently. Even if you're not rich, you've certainly internalized the pervasive elitism that has persisted in mathematics for hundreds of years.
I only think that you must help us to help you. Your questions are all way too vague to answer. Hence we need a dialogue with you to specify your questions. There is unfortunately no shortcut to linear algebra. One has to start at the beginning.
Square matrices are closed under addition...
Wrong. Additive groups are closed under addition. Vector spaces consist of additive groups. Certain sets of matrices, square or not, are closed under addition. A matrix alone cannot be closed, a set of them can be.
... and their own form of multiplication, ...
Wrong. Some sets do, and others don't. It depends on the set. Cp. post #5 above.

but in general do not commute.
Right. Multiplication of arbitrary square matrices is not commutative. A set of diagonal matrices is commutative. Diagonal matrices (over a field) commute.
What algebraic structure then describes this, ...
What is "this"? Lie algebras describe to some extent the degree of commutativity if it was that what you have asked. Every associative algebra ##\mathcal{A}## turns into a Lie algebra via the multiplication
$$[X,Y]:=XY-YX\text{ for all } X,Y\in \mathcal{A}$$
The result ##[X,Y]## is thus one way to describe the failure of commutativity. If we have multiplicative groups instead of associative algebras, then their central series would be appropriate to investigate.
... along with polynomials of matrices and allows us ...
What do polynomials of matrices allow? And again, which set of matrices? Which set of polynomials?
... to amend with other operations,...
Which other operations? There cannot be even an attempt to answer that. This is an invitation for an endless, and meaningless discussion.
... such as differentiation or integration ...
That would be multivariate calculus. A rather broad topic. Do you ask for a book?
... defined on these objects? Would the appropriate algebraic structure be a ring or a field?
You cannot talk about structures if it isn't even clear what the set of such a structure is.

A concatenation of mathematical terms does not make a valid question.

topsquark
fresh_42 said:
I only think that you must help us to help you. Your questions are all way too vague to answer. Hence we need a dialogue with you to specify your questions. There is unfortunately no shortcut to linear algebra. One has to start at the beginning.

Wrong. Additive groups are closed under addition. Vector spaces consist of additive groups. Certain sets of matrices, square or not, are closed under addition. A matrix alone cannot be closed, a set of them can be.

Wrong. Some sets do, and others don't. It depends on the set. Cp. post #5 above.

Right. Multiplication of arbitrary square matrices is not commutative. A set of diagonal matrices is commutative. Diagonal matrices (over a field) commute.

What is "this"? Lie algebras describe to some extent the degree of commutativity if it was that what you have asked. Every associative algebra ##\mathcal{A}## turns into a Lie algebra via the multiplication
$$[X,Y]:=XY-YX\text{ for all } X,Y\in \mathcal{A}$$
The result ##[X,Y]## is thus one way to describe the failure of commutativity. If we have multiplicative groups instead of associative algebras, then their central series would be appropriate to investigate.

What do polynomials of matrices allow? And again, which set of matrices? Which set of polynomials?

Which other operations? There cannot be even an attempt to answer that. This is an invitation for an endless, and meaningless discussion.

That would be multivariate calculus. A rather broad topic. Do you ask for a book?

You cannot talk about structures if it isn't even clear what the set of such a structure is.

A concatenation of mathematical terms does not make a valid question.
Post #2 by WWGD answered the bulk of my inquiry in only a few sentences. Perhaps in all the time you've been buried under nothing but math, you've lost the basic social skills to communicate with laymen, also not uncommon in the world of math.

topsquark, berkeman and PeroK

topsquark
After moderator discussion, the thread has been reopened.

topsquark
I'm almost always interested in learning new subjects, why are you being elitist?
I am not. I try to understand. I basically asked in post #5 about details, you did not answer, I asked you privately per PM without an answer, so I tried a provocative statement:
fresh_42 said:
I don't understand how @topsquark comes into play here, but I think that @askmathquestions isn't really interested in learning a certain subject.
This is clearly a personal opinion. It is my impression. Correct me if I'm wrong. It was by no means offensive. I didn't even say that you weren't willing to learn. I said I don't see a certain subject you are interested in. This was meant as a hidden question about which subject. Again, a provocation? Yes, but not an insult.

Post #2 by WWGD answered the bulk of my inquiry in only a few sentences.
No, he did not. He made an assumption, namely that you would have meant all possible matrices over a ring, and then answered to that assumption. Without such guesses, he wouldn't have been able to give a qualified answer which he did, under his assumption. Without such a condition, we would be in the situation of post #5. You might have felt that he gave you an answer, but I doubt - again my personal opinion - that you understood what he wrote.

topsquark
Post #2 by WWGD answered the bulk of my inquiry in only a few sentences. Perhaps in all the time you've been buried under nothing but math, you've lost the basic social skills to communicate with laymen, also not uncommon in the world of math.
You are asking a math question in a math forum. It would be reasonable to expect "nothing but math".

Communication is a two way street, you don't have to reply to the people trying to help if you aren't happy with how they provide guidance. You are getting expert advice here for free, they are trying to help as best they can; it's the only reason they would respond. If you are too critical you may get what they owe you... nothing.

PeterDonis and topsquark
Post #2 by WWGD answered the bulk of my inquiry in only a few sentences. Perhaps in all the time you've been buried under nothing but math, you've lost the basic social skills to communicate with laymen, also not uncommon in the world of math.
Consider that you are asking a question about rings. Isn't it fair to assume that you have at least some knowledge of what a ring is? So why are you surprised that someone is responding to you with language based on ring theory? You should expect that! If you don't know the theory then you need to say that up front.

-Dan

malawi_glenn
fresh_42 said:
I am not. I try to understand. I basically asked in post #5 about details, you did not answer, I asked you privately per PM without an answer, so I tried a provocative statement:

This is clearly a personal opinion. It is my impression. Correct me if I'm wrong. It was by no means offensive. I didn't even say that you weren't willing to learn. I said I don't see a certain subject you are interested in. This was meant as a hidden question about which subject. Again, a provocation? Yes, but not an insult.

No, he did not. He made an assumption, namely that you would have meant all possible matrices over a ring, and then answered to that assumption. Without such guesses, he wouldn't have been able to give a qualified answer which he did, under his assumption. Without such a condition, we would be in the situation of post #5. You might have felt that he gave you an answer, but I doubt - again my personal opinion - that you understood what he wrote.

Instead of trying to "provoke" people which is clearly rude, you have a few options:

1 is, don't comment on a question you don't like, go do something else with your life
2 is be patient and understanding, work with the poster.
topsquark said:
Consider that you are asking a question about rings. Isn't it fair to assume that you have at least some knowledge of what a ring is? So why are you surprised that someone is responding to you with language based on ring theory? You should expect that! If you don't know the theory then you need to say that up front.

-Dan

I think that's irrelevant to trying to manipulating someone with provocations, though I agree it would be helpful to say that.

Instead of trying to "provoke" people which is clearly rude, you have a few options:

1 is, don't comment on a question you don't like, go do something else with your life
2 is be patient and understanding, work with the poster.

To provoke an answer wasn't my first choice. It was my last. You did not respond to direct questions.

Well, I have a third option: I will strictly apply our rules.

fresh_42 said:
To provoke an answer wasn't my first choice. It was my last. You did not respond to direct questions.

Well, I have a third option: I will strictly apply our rules.
You're saying that as if you "had" or "needed" to do anything, even though I doubt anyone tried to force you to do anything.