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Suppose that there are some matrices. Each matrix in the set must commute with another in the set.

What are the mandatory conditions for this?

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Use "set" instead of "group" in order to avoid any misunderstanding! – DonAntonio Oct 26 '12 at 19:01
Do you really mean "each matrix in the set must commute with every other matrix in the set"? The way you wrote it, you might have only matrix $1$ commuting with matrix $2$, matrix $3$ with matrix $4$, etc. – Robert Israel Oct 26 '12 at 19:51

Over an algebraically closed field, if a family of matrices commute pairwise then they are simultaneously triangularizable. As Robert Israel pointed out in the comments, the converse is not true.

This generalizes to diagonalizable matrices, i.e. a family of diagonalizable matrices commute pairwise if and only if they are simultaneous diagonalizable. This condition is if and only if.

These are the most common results on commuting matrices.

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Not true. The triangular matrices $$ \pmatrix{1 & 1 & 0\cr 0 & 1 & 0\cr 0 & 0 & 1\cr}, \ \pmatrix{1 & 0 & 0\cr 0 & 1 & 1\cr 0 & 0 & 1\cr}$$ don't commute. – Robert Israel Oct 26 '12 at 19:54
@RobertIsrael Yes, I was mistaken. The triangularizabiltiy is only one directional, i.e. commuting matrices are simultaneously triangularizable but not necessarily the other way around. – EuYu Oct 26 '12 at 20:12

There is a quirky condition that does this. If every matrix in the set is a polynomial or, indeed, analytic function of some matrix $A$ (that need not be in the set), they all commute, since a power of $A$ commutes with another power of $A.$ Over the reals this includes items such as $e^A.$

I think it unlikely that one can find such an $A$ for every set for which each pair of matrices commute. It's just a cute idea.

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Community pinged you here :) I would think that it is a completion question - a set is complete if any matrix from the set cannot create any other matrix outside the set by any analytic function. So of course an incomplete set would have such an $A$ outside the set that still commutes with all those in the set. – adam W Jan 11 '13 at 1:15
I've seen this kind of statements, about analytic functions of matrices, in several places on this site and I don't understand them. Every matrix satisfies a polynomial relation (by Cayley-Hemilton if you like) therefore analytic implies polynomial (of degree less than the size of the matrices). – Marc van Leeuwen Feb 12 '13 at 20:51
@MarcvanLeeuwen, yes, now that you mention it. I wouldn't depend on that with an undergraduate asking. PSG are pretty good. – Will Jagy Feb 12 '13 at 23:21

I'll try to rephrase the answer by Will Jagy a bit more explicitly, and add some more detail.

Characterising all sets of commuting matrices (other than by the condition that they commute) won't be easy, because given any set of commuting matrices (and there are certainly such sets that are infinite), than any subset of it will also be a set of commuting matrices. So it is more fruitful to ask for maximal commuting sets of matrices, sets for which nothing outside the set commutes with all of them (for one could then add such a matrix).

Now given any set of commuting matrices, we can always take scalar multiples of one of them, or sums of several of them, to get other matrices that commute with all of them; therefore a maximal set of commuting matrices must be a subspace of the vector space of all matrices (i.e., it is closed under linear combinations). Moreover, it must also be closed under products of matrices (the technical term is that it must be a subalgebra of the algebra of all square matrices). Given any one matrix $A$, the smallest subalgebra that contains $A$ is the set of polynomials in $A$, the linear combinations of powers $A^i$, including $A^0=I_n$.

Correction I wrote here earlier that every maximal commutative subalgebra has dimension $n$ and is the set of polynomials in one matrix (I thought that this was implied by the answer by Will Jagy, but it clearly isn't, and I thought that I could give a somewhat complicated argument for it, but I can't). This is not true. Indeed, there is a $5$-dimensional commutative subalgebra of $M_4(K)$ of matrices of the form $$ \begin{pmatrix}x&0&a&b\\0&x&c&d\\0&0&x&0\\0&0&0&x\end{pmatrix} $$ and for dimension reasons it cannot be the set of polynomials of any one matrix.

A question to which I do not know the answer is can two commuting matrices $X,Y$ always be expressed as polynomials of some third matrix $A$? I initially thought, inspired by the above, that $$ X=\begin{pmatrix}0&0&1&0\\0&0&0&1\\0&0&0&0\\0&0&0&0\end{pmatrix},\qquad Y=\begin{pmatrix}0&0&1&0\\0&0&0&-1\\0&0&0&0\\0&0&0&0\end{pmatrix}, $$ provided a negative answer; however it turns out that for $$ A=\begin{pmatrix}1&0&1&0\\0&0&0&1\\0&0&1&0\\0&0&0&0\end{pmatrix} $$ (which is not in the commutative subalgebra above) one has $Y=A(A-I)=A^2-A$ and $X=A(A-I)(2A-I)=2A^3-3A^2+A$.

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Marc, do you know of a reference for the fact in the last paragraph? – Andrés E. Caicedo Mar 7 '13 at 7:13
@AndresCaicedo I just found out that it is not a fact but a blatant lie, see the answer here. I'll have to correct my answer. – Marc van Leeuwen Mar 7 '13 at 9:35

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