Timeline for Almost commuting matrices, one a projection, is there a nearby projection that commutes?
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 5, 2021 at 21:04 | history | edited | William Bell | CC BY-SA 4.0 |
added 52 characters in body
|
| Mar 5, 2021 at 8:16 | comment | added | Mikael de la Salle | @WilliamBell Oops, I meant $P = \begin{pmatrix} 1 &0\\0&0\end{pmatrix}$, and the conclusion is $\|P-Q\| \geq \frac{1}{\sqrt{2}}$. Sorry for the confusion. | |
| Mar 5, 2021 at 0:07 | history | edited | William Bell | CC BY-SA 4.0 |
Counterexample
|
| Mar 5, 2021 at 0:03 | comment | added | William Bell | Perhaps you meant something else, but I found a simple counterexample soon after this. | |
| Mar 4, 2021 at 23:48 | comment | added | William Bell | @MikaeldelaSalle I am not sure I understand your example, isn't P a projection near P that commutes with A? (Since P is the identity in your example) | |
| Mar 4, 2021 at 9:06 | comment | added | LSpice |
TeX note: $\lvert\lvert A\rvert\rvert$ \lvert\lvert A\rvert\rvert spaces badly; prefer $\lVert A\rVert$ \lVert A\rVert. I have edited accordingly.
|
|
| Mar 4, 2021 at 9:04 | history | edited | LSpice | CC BY-SA 4.0 |
\vert\vert -> \Vert
|
| Mar 4, 2021 at 8:44 | comment | added | Mikael de la Salle | For an explicit example, take $P=\begin{pmatrix} 1&0\\0&1\end{pmatrix}$ and $A=\begin{pmatrix} 0&\delta\\ \delta&0\end{pmatrix}$, so that $\|PQ-QP\| \geq \frac{1}{\sqrt{2}}$ for every projection $Q$ that commutes with $A$. | |
| Mar 4, 2021 at 8:40 | comment | added | Mikael de la Salle | If you do not impose any normalization, what you ask is even false in fixed dimension $n=2$. If you fix $A$ and $P$ that do not commute, say with $A$ of norm $<1$, then for $A'=\frac{\delta}{2} A$, you have $\|A' P - P'A\| < \delta$, but any projection that commutes with $A'$ must commute with $A$ and therefore be at distance $\geq c(A,P)$ from $P$. | |
| Mar 4, 2021 at 7:56 | vote | accept | William Bell | ||
| Mar 4, 2021 at 23:44 | |||||
| Mar 4, 2021 at 7:44 | answer | added | Federico Poloni | timeline score: 3 | |
| Mar 4, 2021 at 7:41 | comment | added | Federico Poloni | @WilliamBell No, just a scalar. $A = J + \delta e_1 e_n^T$. I went over this in my head and now I am even convinced that it works, with $P = e_1 e_1^T$. Then $\operatorname{Im} Q$ must be a real eigenvector of $A$, and the only one is at a distnace $O(\delta^{1/n})$ from $A$. Taking $n$ sufficiently large gives a counterexample. I will write it as an answer. | |
| Mar 4, 2021 at 7:30 | comment | added | William Bell | @FedericoPoloni Thanks for the suggestions, just checking, is $A(n,1)$ a rank-1 matrix? | |
| Mar 4, 2021 at 7:25 | comment | added | Federico Poloni | Just a hunch, but I would look for counterexamples with badly separated invariant subspaces, such as $A = \begin{bmatrix}0 & 1\\ \delta & 0\end{bmatrix}$ or larger Jordan blocks plus a perturbation in $A(n,1)$. | |
| Mar 4, 2021 at 7:09 | history | edited | William Bell | CC BY-SA 4.0 |
Clarifying for a comment.
|
| Mar 4, 2021 at 7:08 | comment | added | William Bell | @dodd I've tried to clarify a couple things with an edit, but I don't think I understand your remark. | |
| Mar 4, 2021 at 7:00 | comment | added | markvs | What if $n=2$, $Q=\left(\begin{array}{ll}1& 0\\ 0 &0\end{array}\right)$? | |
| Mar 4, 2021 at 5:06 | history | edited | William Bell | CC BY-SA 4.0 |
added 60 characters in body
|
| Mar 4, 2021 at 4:56 | history | asked | William Bell | CC BY-SA 4.0 |