# How does $\sigma(T)$ change with respect to $T$?

Consider $\sigma$ as a mapping which maps $T\in\mathcal{L}(X)$ to $\sigma(T)$, the spectrum of $T$, a compact set in the complex plane.

I wonder whether there is some result concerning how $\sigma(T)$ changes when $T$ changes. For instance, how is the Hausdorff distance between $\sigma(T)$ and $\sigma(S)$ related to $\|T-S\|$? Or something like this.

Thanks!

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This reference may help matwbn.icm.edu.pl/ksiazki/zm/zm22/zm2218.pdf, but it doesn't answer your question for arbitrary bounded operators. – Norbert May 30 '12 at 17:28
@Norbert Thanks for the reference! It's helpful! – Hui Yu May 30 '12 at 18:47

Without assuming additional properties on $S$ and $T$, such as commutation or self-adjointness (on a Hilbert space) — see points 2. and 3. at the end of the answer — continuity of the spectrum as a map from $\mathcal{L}(X)$ to the compact subsets of $\mathbb{C}$ with the Hausdorff distance $d_H$ doesn't hold, because the spectrum can “collapse” under arbitrarily small perturbations, as illustrated by the following simple example:

Let $X = \ell^p(\mathbb{Z})$ with basis $(e_n)_{n \in \mathbb{Z}}$ and $1 \leq p \leq \infty$. Define the operator $S: X \to X$ by $$S(e_n) = e_{n-1}\text{ if }n\neq 0 \quad\text{and}\quad S(e_0) = 0.$$ It is not difficult to check that $\sigma(S) = \overline{\mathbb{D}} = \{\lambda \in \mathbb{C}\,:\,\lvert \lambda\rvert \leq 1\}$: Indeed, $\lVert S \rVert = 1$, so we certainly have the inclusion $\sigma(S) \subset \overline{\mathbb{D}}$ and if $\lvert\lambda\rvert \lt 1$, the vector $v = \sum_{n=0}^\infty \lambda^n e_n$ is an eigenvector of $S$ with eigenvalue $\lambda$, so $\lambda \in \sigma(S)$. It follows from compactness of $\sigma(S)$ that $\sigma(S) \supset \overline{\mathbb{D}}$.

Let now $C$ be the rank one operator defined by $Ce_0 = e_{-1}$ and $C(e_n) = 0$ for $n \neq 0$. Putting $T_\varepsilon = S + \varepsilon C$ we have an invertible operator for $\varepsilon \neq 0$ and $\lVert T_{\varepsilon} - S\rVert = \varepsilon$.

For the spectral radius of $T_\varepsilon$ we have $r(T_{\varepsilon}) =1$, so $\sigma(T_{\varepsilon}) \subset \overline{\mathbb{D}}$. The inverse of $T_{\varepsilon}$ also has spectral radius $r(T_{\varepsilon}^{-1}) = 1$, so $\sigma(T_{\varepsilon}) \subset \partial\mathbb{D}$.

It follows from $\sigma(S) = \overline{\mathbb{D}}$ and $\sigma(T_\varepsilon) \subset \partial \mathbb{D}$ that the Hausdorff distance between $\sigma(S)$ and $\sigma(T_{\varepsilon})$ is at least $1$, while $\lVert S-T_{\varepsilon}\rVert = \varepsilon$ is as small as we wish.

Added: On the other hand, in some sense this is the worst that can happen: T. Kato, Perturbation Theory for Linear Operators, Springer Classics in Mathematics, 1995, proves in §3, section 1. of chapter IV on pp208f the following:

1. The spectrum is upper semicontinuous as a function from $\mathcal{L}(X)$ to the compact subsets of $\mathbb{C}$ with respect to the Hausdorff distance (Remark 3.3).

2. If $S$ and $C$ commute then there is the estimate $d_{H}(\sigma(S),\sigma(S+C)) \leq r(C)$ on the Hausdorff distance of the spectra (Theorem 3.6).

3. Much later: To address a question posed in a comment: if $S$ and $T$ are self-adjoint operators on a Hilbert space then $d_H(\sigma(S),\sigma(T)) \leq r(S-T) = \lVert S - T \rVert$ holds as well, mainly because the resolvent satisfies $\lVert R_T(\lambda)\rVert = 1/d(\lambda,\sigma(T))$, which allows us to apply a variant of the Neumann series, see Kato, Theorem V.4.10, page 291 and section II.1.3, in particular formula (1.13) on page 67. Far more precise results can be found at the beginning of chapter VIII, especially §1.2.

The example above (which I learned from Edi Zehnder) appears as Example 3.8 on page 210.

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This (illuminating) example somewhat reminds me of the ill-conditioning of the Jordan form in numerical linear algebra. – Giuseppe Negro May 30 '12 at 18:03
An illuminating example! But how would the property, say, commutation, change the situation? – Hui Yu May 30 '12 at 18:47
@HuiYu: See edit.${}{}{}$ – t.b. May 30 '12 at 19:38
@t.b. Out of curiosity, why would the result be true for self-adjoint operators? How would one prove, say, that if $A_n \to A$ are (non-commuting and) self-adjoint then $d_H(\sigma(A_n), \sigma(A)) \to 0$? – user12014 Jun 12 '12 at 23:48
@PZZ: you can prove that for s.a. operators you have $d_H(\sigma(A_n),\sigma(A)) \leq r(A_n-A)=\|A_n-A\|$ because the norm of the resolvent of a s.a. operator $B$ is $\|R_B(\lambda)\| = 1/d(\lambda,\sigma(B))$ which allows you to use a variant of the Neumann series. See Kato, Thm V.4.10 on p.291 and also section II.1.3, especially formula (1.13) on p.67. Much more precise results are in chapter VIII, especially 1.2. – t.b. Jun 14 '12 at 0:47