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Endow $\mathbb{C}^{d \times d}$ with the norm induced by the Euclidean norm on $\mathbb{C}^d$. It is well-known (to those who know it well, I guess) that the spectrum $\sigma(A)$ of a matrix $A \in \mathbb{C}^{d \times d}$ depends continuously on $A$ with respect to the Hausdorff distance $\operatorname{d}_{\operatorname{Haus}}$ between spectra.

Actually one can even show that the spectrum is locally Hölder-continuous with exponent $1/d$:

Theorem. Let $A,B \in \mathbb{C}^{d \times d}$ such that $\lVert A-B \rVert^{1/d} \le \frac{1}{2}$. Then $$ \operatorname{d}_{\operatorname{Haus}}(\sigma(A), \sigma(B)) \le \lVert A-B \rVert^{1/d} \max\{1, 4 \lVert A \rVert, 4 \lVert B \rVert \} $$ One can prove this by messing around with resolvent estimates in finite dimensions that are based on the estimate $\lVert A^{-1} \rVert \le \lVert A \rVert^{d-1} / \, \lvert \det A \rvert^d$ for invertible $A$, which can be shown by using the polar decomposition of $A$ and which apparently goes back to [Kato 1960, Lemma 1 in the Appendix].

(I computed the explicit estimate in the theorem for a course in spectral theory that I'm currently teaching, when I unexpectedly entered a "let's do an explicit estimate" mood while preparing the course.)

Remarks:

  • I suspect that messing around with Jordan blocks will show that the exponent $1/d$ for the local Hölder continuity is optimal, but I haven't checked in detail, yet.

    [Edit: As pointed out by Terry Tao in a comment, perturbing a Jordan block does indeed show that the $1/d$ is optimal.]

  • The estimate in the theorem is already a bit simplified: the constant $4$ that occurs on the right hand side can be improved to a number close to $2$ if $A$ and $B$ are sufficiently close.

  • One can also prove a (slightly more involved) estimate under the weaker assumption $\lVert A-B \rVert^{1/d} < 1$; however, the right hand side that I was able to obtain in this case explodes for $\lVert A-B \rVert^{1/d} \to 1$, so that estimate becomes meaningless for $\lVert A-B \rVert^{1/d} \to 1$.

    Hence, I have no idea how a good global estimate would look like.

    I also don't know whether $\lVert A \rVert$ and $\lVert B \rVert$ do really need to appear with exponent $1$ on the right hand side.

  • So I was trying to find such estimates in the literature. I checked Kato's obligatory book on perturbations of linear operators, and also searched a while on the internet, but I could not even locate a reference where the local Hölder continuity with exponent $1/d$ is proved (although this property is probably not surprising, if one considers the resolvent growth close to the eigenvalue for a Jordan block of size $d \times d$).

Question: What is an "optimal" (in some reasonable sense of the word) estimate for $\operatorname{d}_{\operatorname{Haus}}(\sigma(A), \sigma(B))$ in terms of the quantities $\lVert A-B \rVert^{1/d}$, $\lVert A \rVert$, and $\lVert B \rVert$?

I am hoping for an estimate that gives the "right" behaviour for small and large values (and combinations thereof) of $\lVert A-B \rVert^{1/d}$, $\lVert A \rVert$, and $\lVert B \rVert$, i.e., I would be interested in a global rather than a local estimate (as indicated in the third bullet point above).

Sub-question: What is a good reference where I can find detailed information on such quantitative estimates?

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    $\begingroup$ Perturbing the shift matrix (i.e., a Jordan block) does indeed show that the 1/d exponent is sharp, and would probably answer your first question. See e.g., my blog post terrytao.wordpress.com/2008/10/28/when-are-eigenvalues-stable $\endgroup$
    – Terry Tao
    Commented Nov 3, 2022 at 23:58
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    $\begingroup$ @TerryTao: Great, thanks for your comment! Indeed, the matrix $U+E$ (notation from your blog post) is what I had in mind when I wrote "messing around with Jordan blocks" - but I didn't notice that it's so easy to compute the characteristic polynomial of $U+E$. Yet, I don't see whether it really answers my first question, which was rather intended to refer to the third bullet point before the question: i.e., I would be interested in a global estimate, since the estimate in the theorem only holds locally (I think I should have made this a bit clearer in the post; I'll edit accordingly). $\endgroup$ Commented Nov 4, 2022 at 0:29
  • $\begingroup$ This question can be generalized to the quaternionic case since $M_d(\mathbb{C})$ embeds inside $M_d(\mathbb{H})$. Say that a $2d\times 2d$-complex matrix $A$ is a quaternionic matrix if we write $A$ as a block matrix with $2\times 2$ blocks, and each block is of the form $$\begin{bmatrix} w & z \\ -\overline{z} & \overline{w} \end{bmatrix}.$$ Let $M_d(\mathbb{H})$ denote the set of all quaternionic matrices. I suspect that the modulus of continuity of $\sigma$ on $M_d(\mathbb{H})$ is still $\Theta(x^{1/d})$. $\endgroup$ Commented Nov 23, 2022 at 17:57

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