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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