# Approximating commuting matrices by commuting diagonalizable matrices

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