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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.

  • 0
    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 $\mathbb{R}$).2012-11-02
  • 0
    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.2012-11-02
  • 0
    Ah, I see. Not sure then.2012-11-02
  • 2
    Answered here: http://mathoverflow.net/questions/111581/approximating-commuting-matrices-by-commuting-diagonalizable-matrices2012-11-07

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