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.

P.S. I asked this question on math.SE a couple of days ago.

  • 3
    $\begingroup$ The corresponding theorem for general semisimple (I think) Lie algebras is due to Richardson '79. $\endgroup$ – Allen Knutson Nov 6 '12 at 7:37

Over the complex numbers the answer is yes. It was proved by Gerstenhaber that the variety of pairs of commuting matrices is irreducible. This variety has an open subset the set of pairs $(A,B)$ such that $A$ and $B$ are commuting diagonalizable matrices.

Edit: 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.

Further edit for real case: It is sufficient to prove that if $A$ and $B$ are commuting real matrices then there exist matrices $E$ and $F$ such that for all sufficiently small non-zero $\epsilon$, the matrices $A + \epsilon E$ and $B + \epsilon F$ commute and are diagonalizable. This can be done using some details from the proof by Motzkin and Taussky.

Outline: Since $A$ and $B$ commute they preserve each others primary decompositions. This reduces to the case where $A$ and $B$ are primary matrices. I'll deal here with the case where neither $A$ nor $B$ have real eigenvalues. By adding a multiple of the identity matrix to $A$ and $B$ we can assume that $A$ has characteristic polynomial $(x^2 + \theta^2)^m$ and $B$ has characteristic polynomial $(x^2 + \phi^2)^m$. There is a decomposition

$$ V = \left< u_1, v_1 \right> \oplus \cdots \oplus \left< u_m, v_m \right> $$

and matrices $S$, $T$, $M$ and $N$ such that

  • $S$, $T$, $M$, $N$ commute,
  • $A = S(I+M)$ and $B = T(I+N)$,
  • $S$ and $T$ preserve each $2$-dimensional summand $\langle u_i,v_i \rangle$, acting as the matrices $$ \left( \begin{matrix} \cos \theta & \sin \theta \newline -\sin \theta & \cos \theta \end{matrix} \right) \quad\text{and}\quad \left( \begin{matrix} \cos \phi & \sin \phi \newline -\sin \phi & \cos \phi \end{matrix} \right) $$ respectively.
  • $M$ and $N$ are nilpotent and $u_i M \in \langle u_{i+1}, v_{i+1}, \ldots, u_m , v_m \rangle$ and similarly for $v_i M$. (The action of $N$ could be more complicated.)

If $A$ is cyclic then there exists $f \in \mathbf{R}[x]$ such that $B = f(A)$. If $E$ is such that $A + \epsilon E$ is diagonalizable for all sufficiently small non-zero $\epsilon$ then we can define the required perturbation $F$ by $B + \epsilon F = f(A+ \epsilon E)$.

If $A$ is not cyclic then there exists $r > 1$ such that the projection $E_r$ with image $\langle u_r, v_r, \ldots, u_m, v_m \rangle$ and kernel $\langle u_1, v_1, \ldots, u_{r-1}, v_{r-1} \rangle$ commutes with $A$. Then $T(I + N + \epsilon E_r)$ commutes with $A$. Moreover, $E_r$ commutes with $T$, and since $N$ and $M$ commute, the eigenvalues of $N + \epsilon E_r$ are either $0$ or $\epsilon$. The characteristic polynomial of $T(I+N + \epsilon E_r)$ splits into a product of powers of the distinct irreducibles $x^2 + \phi^2$ and $x^2 + \phi^2 - \epsilon$. So we get a further decomposition of $\mathbf{R}^{2m}$, still preserved by $A$, and the result follows by induction on the dimension.


This result was first proved by Motzkin and Taussky (essentially in all characteristics) depending upon how you state the result: the set of pairs of n x n matrices (A,B) where AB = BA and A and B have distinct eigenvalues is Zariski dense in the set of commuting pairs of matrices (which is equivalent to the irreducibility of the commuting variety; both results are in Motzkin-Taussky). The reference is: Pairs of matrices with property L. II, Trans. AMS 80 (1955), 387-401. A trivial consequence of this is that any pair of commuting n x n matrices generates an algebra of dimension at most n. Gerstenhaber proved a slightly stronger result: any pair of commuting n x n matrices is contained in a commutative n-dimensional subalgebra. See Guralnick, LAMA 31 (1992), 71-75 for a slightly easier proof and also some remarks on commuting 3-tuples (this variety is reducible for n at least 28, irreducible for n at most 10). So one cannot in general approximate 3 commuting matrices by commuting semismple matrices. See also Guralnick-Sethuram (Commuting pairs and triples of matrices and related varieties. Linear Algebra Appl. 310 (2000), 139–148) showing that Gerstenhaber's stronger result also follows easily from Motzkin-Taussky.


Here are two references.

  1. K. C. O'Meara and C. I. Vinsonhaler, On approximately simultaneously diagonalizable matrices, Linear Algebra Appl., 412 (2006), 39 - 74.

  2. K. C. O'Meara, J. Clark and C. I. Vinsonhaler, Advanced Topics in Linear Algebra : Weaving Matrix Problems through the Weyr Form, Oxford University Press, 2011.

The seventh chapter of this book discusses this problem. By the way, was it proved by Gerstenhaber or Motzkin & Olga Taussky ?

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    $\begingroup$ It was proved by Motzkin and Taussky. $\endgroup$ – Robert Guralnick Dec 7 '12 at 18:27

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