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Suppose the matrices $A$ and $B$ commute. Do there exists sequences $A_n$ and $B_n$ of matrices such that

  1. $A_n \rightarrow A$, $B_n \rightarrow B$.

  2. Each $A_n$ is diagonalizable and the same for each $B_n$.

  3. For every $n$, $A_n$ commutes with $B_n$.

Moreover, it would be nice if the following property was additionally satisfied: if $A,B$ are real, then $A_n,B_n$ can be chosen to be real as well.

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No, if $A_n \rightarrow A$ then the eigenvalues of $A_n$ will approach the eigenvalues of $A$. If $A$ has nonreal eigenvalues then for large $n$ so will $A_n$, and so cannot be diagonalizable (over $\mathbb{R}$). –  Jair Taylor Nov 2 '12 at 23:12
When I say that a real matrix $A$ is diagonalizable, I mean that $A=VDV^{-1}$, for some $V,D$ with possibly complex entries. Consequently a matrix with complex eigenvalues may be diagonlizable. –  robinson Nov 2 '12 at 23:17
Ah, I see. Not sure then. –  Jair Taylor Nov 2 '12 at 23:19
Answered here: mathoverflow.net/questions/111581/… –  robinson Nov 7 '12 at 21:23
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1 Answer 1

To get this off the Unanswered list:

The question was subsequently posted to MathOverflow, where it was answered in the affirmative. Some useful references from Mark Wildon’s answer:

I'm fairly sure M. Gerstenhaber was the first to prove that the irreducibility result. His paper is: On dominance and varieties of commuting matrices, Annals Math. 73 (1961), 324-348. However the result asked for in the question was already known from Theorem 5 in T. S. Motzkin, Olga Taussky, Pairs of matrices with property L II, Trans. Amer. Math. Soc. 80 (1955), 387-401. There is a short account of this work after Remark 3.4 in the paper by Meara and Vinsonhaler mentioned in SJ's answer.

The last reference appears to be to K. C. O'Meara and C. I. Vinsonhaler, On approximately simultaneously diagonalizable matrices, Linear Algebra Appl., 412 (2006), 39 - 74.

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