Denote by $\( {\mathcal{L}\mathopen{}\left( E\right)\mathclose{}}^{-1} \)$ the set of invertible operators in $\( \mathcal{L}\mathopen{}\left( E\right)\mathclose{} \)$.

If $\( A\in \mathcal{L}\mathopen{}\left( E\right)\mathclose{} \)$ and $\( \mathopen{}\left\lVert{}\mathrm{I}-A\right\rVert\mathclose{}\lt 1 \)$, then $\( A\in {\mathcal{L}\mathopen{}\left( E\right)\mathclose{}}^{-1} \)$, and $\( \mathopen{}\left\lVert{}\mathrm{I}-{A}^{-1}\right\rVert\mathclose{}\leq \frac{\mathopen{}\left\lVert{}\mathrm{I}-A\right\rVert\mathclose{}}{\mathopen{}\left(1-\mathopen{}\left\lVert{}\mathrm{I}-A\right\rVert\mathclose{}\right)\mathclose{}} \)$.

The set of invertible operators $\( {\mathcal{L}\mathopen{}\left( E\right)\mathclose{}}^{-1} \)$ is an open subset of $\( \mathcal{L}\mathopen{}\left( E\right)\mathclose{} \)$.

The inversion map $\( A\mapsto {A}^{-1} \)$ is continuous from $\( {\mathcal{L}\mathopen{}\left( E\right)\mathclose{}}^{-1} \)$ to itself.

The spectrum of $\( A\in \mathcal{L}\mathopen{}\left( E\right)\mathclose{} \)$ is $$\[ \mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}= \mathopen{}\left\{\, λ\in \mathbb{C}\,\middle\vert\, λ\mathrm{I}-A\notin {\mathcal{L}\mathopen{}\left( E\right)\mathclose{}}^{-1}\,\right\}\mathclose{} \text{.} \]$$

Let $\( E= \mathrm{l}^{0} \)$ and consider the right shift operator $$\[ R\mathopen{}\left( {x}_{1}, {x}_{2}, \dotsc\right)\mathclose{}= \mathopen{}\left(0, {x}_{1}, {x}_{2}, \dotsc\right)\mathclose{}\text{.} \]$$ Suppose $\( R\mathopen{}\left( \mathbf{x}\right)\mathclose{}= λ\mathbf{x} \)$, $\( λ\in \mathbb{C} \)$, $\( \mathbf{x}\in E \)$. Then $\( λ{x}_{1}= 0 \)$, $\( λ{x}_{2}= {x}_{1} \)$, $\( λ{x}_{i}= {x}_{i-1} \)$, which means $\( \mathbf{x}= \mathbf{0} \)$. So, there are no eigenvalues. However, $\( 0\in \mathop{\sigma}\mathopen{}\left( R\right)\mathclose{} \)$ since $\( R \)$ is not invertible. It turns out that $\( \mathop{\sigma}\mathopen{}\left( R\right)\mathclose{}= \mathopen{}\left\{\, z\in \mathbb{C}\,\middle\vert\, \mathopen{}\left\lvert{}z\right\rvert\mathclose{}\leq 1\,\right\}\mathclose{} \)$.

For $\( A\in \mathcal{L}\mathopen{}\left( E\right)\mathclose{} \)$, $\( \mathop{\sigma}\mathopen{}\left( A\right)\mathclose{} \)$ is closed and bounded. Moreover, $\( \mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}\subseteq \mathopen{}\left\{\, λ\in \mathbb{C}\,\middle\vert\, \mathopen{}\left\lvert{}λ\right\rvert\mathclose{}\leq \mathopen{}\left\lVert{}A\right\rVert\mathclose{}\,\right\}\mathclose{} \)$. In particular, $\( \mathop{\sigma}\mathopen{}\left( A\right)\mathclose{} \)$ is compact.

Consider $\( \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathopen{}\left[a, b\right]\mathclose{}\right)\mathclose{} \)$ and $\( f\in \mathrm{C}\mathopen{}\left( \mathopen{}\left[a, b\right]\mathclose{}\right)\mathclose{} \)$. Define $\( \mathrm{M}_{f}\mathopen{}\left( x\mathopen{}\left( t\right)\mathclose{}\right)\mathclose{} \)$ to be $\( f\mathopen{}\left( t\right)\mathclose{}x\mathopen{}\left( t\right)\mathclose{} \)$. We claim $\( \mathop{\sigma}\mathopen{}\left( \mathrm{M}_{f}\right)\mathclose{}= f\mathopen{}\left( \mathopen{}\left[a, b\right]\mathclose{}\right)\mathclose{} \)$. If $\( λ\notin f\mathopen{}\left( \mathopen{}\left[a, b\right]\mathclose{}\right)\mathclose{} \)$, then the continuous function $\( λ-f \)$ has a reciprocal, so $\( λI-\mathrm{M}_{f}= \mathrm{M}_{\mathopen{}\left(λ-f\right)\mathclose{}} \)$ is invertible, that is $\( λ\notin \mathop{\sigma}\mathopen{}\left( \mathrm{M}_{f}\right)\mathclose{} \)$. Conversely, if $\( λ= f\mathopen{}\left( {t}_{0}\right)\mathclose{} \)$ for $\( {t}_{0}\in \mathopen{}\left[a, b\right]\mathclose{} \)$, define $\( {J}_{n}= \mathopen{}\left\{\, t\in \mathopen{}\left[a, b\right]\mathclose{}\,\middle\vert\, , \mathopen{}\left\lvert{}f\mathopen{}\left( t\right)\mathclose{}-λ\right\rvert\mathclose{}\lt \frac{1}{n}, \,\right\}\mathclose{} \)$. Let $\( {δ}_{n} \)$ be the measure of this set. Consider $\( {g}_{n}\mathopen{}\left( t\right)\mathclose{}\in \mathrm{L}^{\mathrm{2}} \)$ given by $$\[ {g}_{n}\mathopen{}\left( t\right)\mathclose{}= \begin{cases}\frac{1}{\sqrt{{δ}_{n}}}, & t\in {J}_{n} ; \\ 0, & t\notin {J}_{n}\text{.} \end{cases} \]$$ This makes $\( \mathopen{}\left\lVert{}{g}_{n}\right\rVert\mathclose{}= 1 \)$. Suppose $\( T= {\mathopen{}\left(λI-\mathrm{M}_{f}\right)\mathclose{}}^{-1} \)$. Then $$\[ \mathopen{}\left(λI-\mathrm{M}_{f}\right)\mathclose{}{g}_{n}= \begin{cases}\frac{λ-f\mathopen{}\left( t\right)\mathclose{}}{\sqrt{{δ}_{n}}}, & t\in {J}_{n} ; \\ 0, & t\notin {J}_{n}\text{.} \end{cases} \]$$ Thus $\( {\mathopen{}\left\lVert{}\mathopen{}\left(λI-\mathrm{M}_{f}\right)\mathclose{}{g}_{n}\right\rVert\mathclose{}}^{2}\leq \frac{μ\mathopen{}\left( {J}_{n}\right)\mathclose{}}{{n}^{2}{δ}_{n}}= \frac{1}{{n}^{2}} \)$. An operator that takes a sequence of unit vectors to 0 cannot have a bounded inverse, so $\( λ\in \mathop{\sigma}\mathopen{}\left( \mathrm{M}_{f}\right)\mathclose{} \)$.

Unless otherwise specified, in what follows, $\( H \)$ is a Hilbert space. Write $\( λ= λI \)$ for $\( λ\in \mathbb{C} \)$.

For $\( T \)$ and $\( S \)$ in $\( \mathcal{L}\mathopen{}\left( H\right)\mathclose{} \)$, we have

1. $\( \mathopen{}\left(TS\right)\mathclose{}^{*}= S^{*} T^{*} \)$.
2. $\( S\in { \mathcal{L}\mathopen{}\left( H\right)\mathclose{} }^{-1} \)$ implies $\( S^{*}\in { \mathcal{L}\mathopen{}\left( H\right)\mathclose{} }^{-1} \)$ with $\( {\mathopen{}\left( S^{*}\right)\mathclose{}}^{-1}= \mathopen{}\left({S}^{-1}\right)\mathclose{}^{*} \)$.
3. $\( \mathop{\sigma}\mathopen{}\left( T^{*}\right)\mathclose{}= \overline{\mathop{\sigma}\mathopen{}\left( T\right)\mathclose{}} \)$.

Let $\( S \)$ be the unilateral shift on $\( \mathrm{l}^{2} \)$. Then $\( \mathop{\sigma}\mathopen{}\left( S\right)\mathclose{} \)$ is the closed unit disc with a proof sketch as follows: For $\( \mathopen{}\left\lvert{}λ\right\rvert\mathclose{}\lt 1 \)$, notice $\( {\mathbf{ξ}}_{λ}= \mathopen{}\left(1, λ, {λ}^{2}, \dotsc\right)\mathclose{}\in \mathrm{l}^{2} \)$ and $\( S^{*}\mathopen{}\left( {\mathbf{ξ}}_{λ}\right)\mathclose{}= \mathopen{}\left(λ, {λ}^{2}, \dotsc\right)\mathclose{}= λ{\mathbf{ξ}}_{λ} \)$. So, the open unit disc is contained in $\( \mathop{\sigma}\mathopen{}\left( S^{*}\right)\mathclose{} \)$. Then $\( \mathop{\sigma}\mathopen{}\left( S^{*}\right)\mathclose{} \)$ is the closed unit disc and so $\( \mathop{\sigma}\mathopen{}\left( S\right)\mathclose{} \)$ is also the closed unit disk.