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Let $A\in\mathbb{R}^{n\times n}$ be diagonalisable (over $\mathbb{C}$) with pairwise distinct eigenvalues $\lambda_1,\ldots,\lambda_n\in\mathbb{C}$, and suppose that

$$\tag{1}S^{-1}\cdot A\cdot S = \mathrm{diag}[\lambda_1,\cdots,\lambda_d]=:\Lambda \qquad \text{for some invertible } S\in\mathbb{C}^{n\times n}.$$

Assume that there is another matrix $T\in\mathbb{R}^{n\times n}$ which 'quasi-diagonalises' $A$ in the sense that

$$\tag{2}T^{-1}\cdot A\cdot T =\Lambda + \Delta \qquad \text{for some} \qquad \Delta\in\mathbb{R}^{n\times n} \ \text{ with } \ \|\Delta\|<\varepsilon$$ where $\varepsilon>0$ is 'small' and $\|\cdot\|$ is a matrix norm of your choice (on $\mathbb{C}^{n\times n}$).

Question: Can we infer that for $\varepsilon>0$ small enough, the matrices $S$ and $T$ are close to one another in the sense that

$$\tag{3}\mathrm{inf}\{\|T\cdot S^{-1} - D\cdot P\| \mid \text{$D\in\mathbb{C}^{n\times n}$ diagonal & invertible}, \ \text{$P$ permutation}\} \ \lesssim \ \|\Delta\|$$

In other words, can we infer that for $\Lambda$ and $\tilde{\Lambda}:=\Lambda + \Delta$ almost identical, the columns of $S$ and $T$ (which define a basis for the almost identical representations $\Lambda$ and $\tilde{\Lambda}$ of the endomorphism $A$) almost coincide up to order and scale?

Any references or hints, or indeed counterexamples, are appreciated.

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    $\begingroup$ This Math.SE post seems relevant: math.stackexchange.com/questions/1133071/… In particular, the standard reference for this sort of thing is "Kato, T., Perturbation theory for linear operators. Springer-Verlag, 1976" $\endgroup$ Feb 3, 2021 at 14:20
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    $\begingroup$ Obviously, this isn't working if $A=1$, so you'll probably need a more quantitative assumption than non-degenerate eigenvalues (a bound on $\min |\lambda_j-\lambda_k|$). $\endgroup$ Feb 3, 2021 at 15:26

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