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Suppose that there are many matrices $A_i\in M_n(C)$, $i=1,2,3,\cdots,m$, that almost commute with each other (what I means is that they are not commute with each other, but $||[A_i, A_j]||$ is very small). Can I slightly change the matrices, e.g., $A_i^\prime:=A_i+\delta_i$ with $||\delta_i||$ very small, such that the new matrices $A_i^\prime$ commute with each other?

user1551
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D. J. Zhang
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  • You said that you have "many" matrices $A_i \in M_n(\mathbb{C})$, with $i=1,\ldots,n$. Did you mean these two $n$'s to be equal? – Pedro M. May 05 '15 at 14:13
  • They are not necessary to be equal. Thank you for pointing this. – D. J. Zhang May 05 '15 at 14:22
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    Then I suggesting changing the second $n$ to $m$, or maybe $N$. Now, as for the question per se, did you try to understand the case of $2 \times 2$ matrices? Say, even for two or three $2 \times 2$ matrices? What can you say in these cases? – Pedro M. May 05 '15 at 18:12
  • Tkanks for your adivise. I can prove that when there are two matrices, one can slightly change them such that the resulting matrices commute with each other. Besides, it is obviously that if the matrices $A_i$, $i=1,2,\cdots,m$, already commute with each other, there is no need to change them. Thus, I conjecture that for any number of matrices that almost commute with each other, one can slightly change them so that they commute with each other. – D. J. Zhang May 06 '15 at 02:27
  • It is very nice if you can prove the case of $2\times2$ matrices. Any comments or ideas are welcomed. – D. J. Zhang May 06 '15 at 02:29

2 Answers2

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If one of these matrices has distinct eigenvalues, assume it is $A_1$, then any small pertubation, $A_1+\delta_1$, has distinct eigenvalues too. Now, any matrix that commutes with a matrix with distinct eigenvalues must be a polynomial of this matrix. So $A_i+\delta_i=p(A_1+\delta_1)$, for some polynomial $p\in\mathbb{C}[x]$.

Claim: $\dim(\text{span}\{A_1,\ldots,A_m\})\leq\dim(\text{span}\{A_1+\delta_1,\ldots,A_m+\delta_m\})$, if $\|\delta_i\|$ is very small for every $i$.

So if $\dim(\text{span}\{A_1,\ldots,A_m\})>\dim(\{p(A_1+\delta_1), p\in\mathbb{C}[x]\})$ then $\dim(\text{span}\{A_1+\delta_1,\ldots,A_m+\delta_m\})>\dim(\{p(A_1+\delta_1), p\in\mathbb{C}[x]\})$ and it is not possible to have $\text{span}\{A_1+\delta_1,\ldots,A_m+\delta_m\}\subset \{p(A_1+\delta_1), p\in\mathbb{C}[x]\}$.

For example: By Calley-Hamilton, $\dim(\{p(A_1+\delta_1), p\in\mathbb{C}[x]\})\leq n$. If $\dim(\text{span}\{A_1,\ldots,A_m\})>n$ and $A_1$ has distinct eigenvalues then the answer for your question is no.

Proof of the Claim: Assume $\dim(\text{span}\{A_1,\ldots,A_m\})=k$ and wlog assume $A_1,A_2,\ldots,A_k$ are linear indepedent.

Consider the continuous funtion $f:\{(z_1,\ldots,z_k),|z_1|+\ldots+|z_k|=1\}\rightarrow\mathbb{R}$, $f(z_1,\ldots,z_k)=\|z_1A_1+\ldots+z_kA_k\|$.

Since the domain is compact and $f(z_1,\ldots,z_k)>0$ then exists $\epsilon>0$ such that $f(z_1,\ldots,z_k)>\epsilon$.

Now, if $a_1(A_1+\delta_1)+\ldots+a_k(A_k+\delta_k)=0$ with $|a_1|+\ldots+|a_k|>0$ then $\epsilon<\frac{1}{|a_1|+\ldots+|a_k|}\|a_1A_1+\ldots+a_kA_k\|=\frac{1}{|a_1|+\ldots+|a_k|}\|a_1\delta_1+\ldots+a_k\delta_k\|\leq\max\{\|\delta_i\|, i=1,\ldots,k\}$.

So if $\max\{\|\delta_i\|, i=1,\ldots,k\}<\epsilon$ then $A_1+\delta_1,\ldots,A_k+\delta_k$ are l.i. $\square$

Daniel
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It doesn't seem possible in general. For instance it is shown in https://www.jstor.org/stable/2047216 that there are concrete $n\times n$ matrices $A$ and $B$ such that $\|[A,B]\|=\|AB-BA\|\leq 2/n$ but for which $\|A-R\|+\|B-S\|\geq 1-1/n$ for all $n\times n$ matrices $R$ and $S$ with $[R,S]=0$.

However, Lin's theorem (for a proof see http://web.math.ku.dk/~rordam/manus/short.pdf) says that if $A$ and $B$ are self-adjoint $n\times n$ matrices that are bounded with norm less than $1$, then one can find nearby commuting self-adjoint matrices.

Condo
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