Pseudospectrum of an $n\times n$ matrix has at most n connected components For $\epsilon&gt;0$, the $\epsilon$ pseudospectrum of an $n\times n$ matrix $A$ is given by $\sigma_{\epsilon}(A)=\{z\in \mathbb{C}:\|(z-A)^{-1}\|&gt;\epsilon^{-1}\}$, with the convention that $\|(z-A)^{-1}\|=\infty$ if $z$ is an eigenvalue of $A$.

It can be shown that $\sigma_{\epsilon}(A)$ is a compact subset of $\mathbb{C}$. It is to be proved that $\sigma_{\epsilon}(A)$ has at most $n$ connected components, each of which contains an eigenvalue of $A$. To prove this, suppose $U$ were a component of $\sigma_{\epsilon}(A)$ that contains no eigenvalue of $A$. We are to use the fact that $\text{log} \|(z-A)^{-1}\|$ is a subharmonic function and thus satisfies the maximum principle. For each $z \in U, \|(z-A)^{-1}\|&gt;\epsilon^{-1}$. How do we use the maximum principle to show that this is possible only if $U$ contains an eigenvalue $\lambda$ for which $\|(\lambda-A)^{-1}\|=\infty$?
If $\sigma_\epsilon(A)$ has $&gt;n$ components each of which contains an eigenvalue of $A$, then the $n\times n$ matrix $A$ would have $&gt;n$ (distinct) eigenvalues.
If $\sigma_\epsilon(A)$ has $&gt;n$ components each of which contains an eigenvalue of $A$, then the $n\times n$ matrix $A$ would have $&gt;n$ (distinct) eigenvalues.