Looking for a counterexample regarding operator norm and spectra

Question

I have proven that if $\mathcal{A}$ is a unital Banach algebra and $r>0$, $x,y$ commuting elements in $\mathcal{A}$ such that $\Vert x - y \Vert < r$ then $\sigma_\mathcal{A}(y) \subseteq B_r (\sigma_\mathcal{A}(x)):=\bigcup_{t\in \sigma_\mathcal{A}(x)}B_r(t).$ What I'm now looking for is a counterexample to show that this does not hold in general for non-commuting elements. However, despite expecting I could find nice and easy counterexamples in $M_n(\mathbb{C})$, I have not been able to find any. Do they exist, or do I need to look at a nastier Banach algebra?

Answer 1

Let $x=\begin{bmatrix} 0 & \frac{1}{\epsilon} \\ 0 & 0 \end{bmatrix}$, $y=x +\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}$. Then $x$ has spectrum $\{ 0 \}$, $y-x$ has norm of order $1$ and $y$ has an eigenvalue of order $\frac{1}{\epsilon}$. So for sufficiently small $\epsilon$ the desired inclusion is violated.

Remark: this behaviour is impossible if $x$ is a normal matrix (or even a normal element of a $C^*$ algebra). This is because the resolvent of $y$ admits a geometric series expansion in powers of $(x-z)^{-1}(y-x)$ and for $x$ normal, $\| (x-z)^{-1} \|$ is bounded by the inverse of distance between the scalar $z$ and the spectrum of $x$. Thus for $\| y-x \| $ smaller than this distance the series converges, and hence $z$ is in the resolvent set.