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I'm working on some beginner group theory, and I understand that there is an issue where most matrices are not commutative (i.e. Abelian). However, I am interested in solving for the properties of matrices that make them Abelian. I was wondering, what is a good way to start exploring the properties of matrices that could make them Abelian? I haven't done much of this "exploratory math" proofs, but I hear it is good practice.

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Eliza
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    What does it mean for a matrix to be commutative? Do you mean a matrix which commutes with every other? – C. Falcon Jan 06 '16 at 22:58
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    It is not so much that a specific matrix has the property, but rather that two matrices together commute. The matrices that commute with all other matrices are quite uninteresting as they are simply scalar multiples of the identity matrix. You can, however, find $A$ and $B$ such that $AB=BA$ with neither being a scalar multiple of the identity. – JMoravitz Jan 06 '16 at 22:59
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    Possible duplicate of A linear operator commuting with all such operators is a scalar multiple of the identity. – JMoravitz Jan 06 '16 at 22:59
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    @JMoravitz Yes, this is most likely what the OP means, but it's a little disheartening to hear the OP excited about exploring her idea only for you to spoil the answer in the comments. :( – Lynn Jan 06 '16 at 23:23
  • If you mean, What are the $n \times n$ matrices $A$ such that $AB = BA $for all $n \times n$ matrices $B$? I'd suggest writing down a general matrix $A$ and looking for conditions on the entries $a_{ij}$ of $A$ such that $A$ commutes with some simple matrices $B$. For purposes of matrix multiplication, the simplest matrices that are not multiplies of $I$ are the matrices $E_{ij}$, which respectively have entry $e_{ij} = 1$ and all other entries zero. – Travis Willse Jan 07 '16 at 00:05
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    It might be interesting for you to consider the conditions on a set of matrices needed for any two matrices in the set to commute with each other. For instance, a set containing the identity matrix and (any) one other matrix has this property. Are there larger sets that work? – mjqxxxx Jan 07 '16 at 04:53

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Provided that you mean by "$A$ an abelian matrix" the following :

$$A\text{ is a matrix such that } BA=AB\text{ for all matrices }B$$

Usually we say that "$A$ is central".

Then it is indeed an interesting exercise. A first approach would be to write down the equations (in the coefficients of $A$) given by $AB=BA$ for some $B$'s nd try to understand what is going on. This takes a lot of time and is not the best approach.

The best approach is to think of matrices as endomorphisms of some vector space. Indeed say you work over a field $K$ and matrices of size $n$.

Define $E:=K^n$, it has a canonical base $e_1,...,e_n$.

If $M$ is a matrix in $M_{n,n}(K)$ then $M$ can be seen as an endomorphism of $E$ by sending $e_i$ to the vector $M.e_i$.

If you have $AB=BA$ then the associated endomorphisms commute as well. Try now to prove the following lemma :

If $E$ is a $K$-vectorial space, $f$ and $g$ are two linear applications from $E$ to $E$. The relation $fg=gf$ implies that any eigenspace of $f$ is stable by $g$.

After you have done this you conclude that $A$ (as an endomorphism) stabilizes every eigenspace of every matrix. Uses this to show that $Vect(e_1)$,...,$Vect(e_n)$ and $Vect(e_1+...+e_n)$ are all stable by $A$ and conclude.

Clément Guérin
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