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How does one find a set of square $n \times n$ matrices that commute with each other?

It seems that there were some stuffs said about invertible and non-invertible, but not about a set of matrices that commute with each other.

Of course, there will be different possible sets, and my question is can every natural number bigger than 2 can be guaranteed to be the cardinality of such set, and if it does, how can we derive such set?

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  • There is a trivial answer to your question: yes, as for every $k$, ${I_n,2I_n,\ldots,kI_n}$ is such a set. And you can get sets like that of cardinaity up to that of you underlying field. Unless I misread again. – Julien Apr 16 '13 at 03:10
  • @julien, so what is the maximum dimension? – Will Jagy Apr 16 '13 at 03:15
  • @WillJagy Not sure I understand. The OP asks about sets and their cardinality. Not dimension. Oh, I see, your question in your answer... – Julien Apr 16 '13 at 03:17
  • @julien, yes, and you have answered that, as I read it. But I find that I do not know this other thing, and perhaps you do. – Will Jagy Apr 16 '13 at 03:18

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BACKGROUND: Given a matrix, is there always another matrix which commutes with it?

Start with any fixed square matrix $A,$ size $n$ by $n$ as you say. Cayley-Hamilton says that $A$ satisfies a polynomial of degree $n.$ Which means that any polynomial in $A$ can be re-written as $$ a_0 I + a_1 A + a_2 A^2 + \cdots + a_{n-1} A^{n-1}. $$ All such matrices commute with each other.

Someone once pointed out to me that this also applies to any real analytic function of $A,$ such as $e^A.$ I think it was Marc van Leeuwen. Anyway, that is correct. So, you get a vector space of matrices of dimension $n,$ out of the full dimension of $n^2.$ That would be my question, is $n$ the maximum dimension of a vector subspace of matrices, all of which commute with each other?

Note, you want $A$ to have $n$ distinct eigenvalues to get full value out of this. For example, if $A=I,$ all polynomials are just $a_0 I,$ dimension 1.

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Will Jagy
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  • @julien, maybe, maybe not. $A^j A^k = A^k A^j.$ – Will Jagy Apr 16 '13 at 02:48
  • Actually, as you can see, I deleted my comment even before that. Sorry about that. – Julien Apr 16 '13 at 02:48
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    @julien, this way, it keeps the reader on his toes. Or her toes. – Will Jagy Apr 16 '13 at 02:50
  • That's a good question you raise, +1. – Julien Apr 16 '13 at 03:25
  • Such a vector space of dimension $k$ generates a commutative subalgebra of dimension $\geq k$. Now it is known for the maximum dimension of a commutative subalgebra in $M_n(K)$ is $1+\lfloor n^2/4\rfloor$. So your $k$ is not greater than that. Given the example in Robin Chapman's answer, this seems to be optimal for subspaces too. – Julien Apr 16 '13 at 03:42
  • @julien, thanks, I'll go through that. So you just get max $=n$ for $n = 2,3.$ Improves for $n \geq 4.$ – Will Jagy Apr 16 '13 at 03:49
  • The set of all diagonal matrices are commuting,and its dimension is n. – R Salimi Apr 16 '13 at 20:14
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    Over $\mathbb{C}$, unless some $A$ in a maximal subspace is similar to a single Jordan block, the dimension of that maximal subspace is always greater than $n$. Suppose $A$ has $n$ repeated eigenvalues. Then WLOG, we may assume that $A=J_1\oplus\cdots\oplus J_m$, where $m>1$ each $J_i$ is a Jordan block. So, $A$ at least commutes with those upper block triangular matrices of the form $B=(B_{ij}){i,j=1}^m$, such that $B$ has conforming partitioning to $A$ and each $B{ij}\ (i\le j)$ is an upper triangular (and perhaps rectangular) circulant matrix. – user1551 Apr 17 '13 at 04:27