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Suppose that $P, A, Q \in \mathbb{M}^{n \times n}(\mathbb{R})$ (I'm still interested if it must be done over $\mathbb{C}$), (EDIT:) suppose that $A$ is given, $P$ is an orthogonal projection, and $\lVert PA-AP \rVert < \delta$, is there some $Q$ an orthogonal projection such that $\Vert P-Q \rVert < \epsilon(\delta)$ (such that as $\delta \to 0$, so does $\epsilon$) and $QA = AQ$? Obviously if there's a broader result than just the finite-dimensional case, then that's good too.

I've seen results vaguely like this, e.g. Lin's Theorem is a result about almost-commuting normal matrices. However I haven't seen a result about projections of this kind.

It would be nice if this were true, I hope to learn more about this type of result, but if the good people here could point me to this result or a similar result from which I could jump off, I'd very much appreciate it.

EDIT: A simple counterexample to my suggestion: $$ A = \begin{bmatrix}1 & \delta \\ 0 & 1\end{bmatrix} $$ $$ P = \begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix} $$ The only non-trivial invariant subspaces are $\mathbb{R}^2$ and the span of $\begin{bmatrix}1 & 0\end{bmatrix}$, but P is not near to the projection for either. But $\lVert PA-AP \rVert = \delta$.

Another counterexample is given in the comments.

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    $\begingroup$ 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 $A(n,1)$. $\endgroup$ – Federico Poloni Mar 4 at 7:25
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    $\begingroup$ @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. $\endgroup$ – Federico Poloni Mar 4 at 7:41
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    $\begingroup$ 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 $P$. $\endgroup$ – Mikael de la Salle Mar 4 at 8:40
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    $\begingroup$ 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 $A$. $\endgroup$ – Mikael de la Salle Mar 4 at 8:44
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    $\begingroup$ @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. $\endgroup$ – Mikael de la Salle Mar 5 at 8:16
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EDIT: I realized that this does not work, because $PA$ has one nonzero too much, sorry. $A$ and $P$ can be simultaneously almost-triangularized, but they don't almost-commute. I'm leaving it up as an attempt, but it doesn't deserve acceptance / upvotes.

Suppose such $\delta<1$ and $\epsilon(\delta)$ exist. Take $A = J + \delta e_1 e_n^T$, $P = e_1 e_1^T$ where $J$ is a nilpotent Jordan block of size $n$, and $e_1,e_n$ are the first and last column of the identity matrix $I$. Since $\|P-Q\|<1$, and orthogonal projection matrices have integer trace equal to their rank, $Q$ must have rank 1. Then $\operatorname{Im} Q$ must be a real eigenvector of $A$, and the only one is $[1\, \delta^{1/n}\, \delta^{2/n}\, \dots \, \delta^{1-1/n}]^T$ (and its multiples). So $Q$ is at a distance $O(\delta^{1/n})$ from $P$. Hence $\epsilon(\delta) > C\delta^{1/n}$ for some $\delta$. This holds for all $n$, and thus $\epsilon(\delta) \geq 1$, contradicting $\epsilon(\delta) \to 0$.

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  • $\begingroup$ This is a good answer to the question as I wrote it, but I don't mind dimension-dependence too much, so do you happen to have a suggestion for if I write $\epsilon(\delta, n)$? $\endgroup$ – William Bell Mar 4 at 7:55
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    $\begingroup$ @WilliamBell Sorry, I have noticed that there is a mistake here, in that $A$ and $P$ do not almost-commute, although they can be almost-simultaneously triangularized. You probably want to un-accept this. :/ $\endgroup$ – Federico Poloni Mar 4 at 8:46
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    $\begingroup$ (On the other hand, I believe that the simpler approach in Mikael de la Salle's comment works) $\endgroup$ – Federico Poloni Mar 4 at 8:48
  • $\begingroup$ Thanks anyways. :) $\endgroup$ – William Bell Mar 4 at 23:47

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