Spectrum and resolvent of a bounded operator

Let $X$ be a complex Banach space (Banach space, The Euclidean metric, convergence, Cauchy sequences, and continuity on the complex plane) and let $T\in \mathcal{B}(X)$ be a bounded linear operator (A bounded linear operator between normed spaces, The spaces (\mathcal B(X,Y)) and (\mathcal B(X)) of bounded linear operators). For a scalar $\lambda \in \mathbb{C}$ write $\lambda I-T$ for the bounded linear operator $x\mapsto \lambda x-Tx$ (Linear map between vector spaces over the same field).

• The resolvent set of $T$ is $\rho (T):=\{\lambda \in \mathbb{C}:\lambda I-T\text{ is bijective and its inverse is a bounded operator }X\to X\}\mathrm{.}$ For $\lambda \in \rho (T)$ the inverse $R(\lambda ,T):=(\lambda I-T{)}^{-1}$, an element of $\mathcal{B}(X)$, is the resolvent operator of $T$ at $\lambda$.
• The spectrum of $T$ is its complement $\sigma (T):=\mathbb{C}\setminus \rho (T)\mathrm{.}$
• A scalar $\lambda$ is an eigenvalue of $T$ when $\ker (\lambda I-T)\mathrm{\ne }\{0\}$; the nonzero vectors of that kernel are the eigenvectors of $T$ for $\lambda$, and $E_{\lambda }(T):=\ker (\lambda I-T)$ is the eigenspace. The set of eigenvalues is the point spectrum of $T$; plainly the point spectrum is contained in $\sigma (T)$, since an operator with nonzero kernel is not injective.
• For $\lambda \in \mathbb{C}$ the generalized eigenspace of $T$ at $\lambda$ is $G_{\lambda }(T):={\bigcup }_{n\ge 1}\ker ((T-\lambda I{)}^n),$ an increasing union of linear subspaces (Linear subspace of a vector space); the union is a linear subspace because the union is increasing. If $\lambda$ is an eigenvalue and $G_{\lambda }(T)$ is finite dimensional, ${\dim }_{\mathbb{C}}{G}_{\lambda }(T)$ (Finite-dimensional vector space, and its dimension ${\dim }_{F}V$; infinite-dimensional means having no finite basis) is the algebraic multiplicity of the eigenvalue $\lambda$.

Every element of $G_{\lambda }(T)\setminus \{0\}$ produces an eigenvector, and conversely. If $(T-\lambda I{)}^{n}x=0$ with $x\mathrm{\ne }0$, let $k\ge 1$ be least with $(T-\lambda I{)}^{k}x=0$; then $y:=(T-\lambda I{)}^{k-1}x$ is nonzero and satisfies $(T-\lambda I)y=0$. So $G_{\lambda }(T)\mathrm{\ne }\{0\}$ exactly when $\lambda$ is an eigenvalue. In particular $\lambda \in \rho (T)$ implies $G_{\lambda }(T)=\{0\}$. If $\lambda$ is an eigenvalue and $G_{\lambda }(T)$ is finite dimensional, then the algebraic multiplicity is defined and is at least ${\dim }_{\mathbb{C}}{E}_{\lambda }(T)$, because $E_{\lambda }(T)\subseteq G_{\lambda }(T)$.

Conventions. Only complex scalars are treated here; the real case is handled by complexification on a later page, so no spectrum is attached here to a bounded operator on a real Banach space. The definition is purely one of vocabulary; it asserts no nonemptiness of $\sigma (T)$, no openness of $\rho (T)$ and no continuity of $\lambda \mapsto R(\lambda ,T)$, all of which are proved separately. Since $0I-T=-T$, the number $0$ lies in $\sigma (T)$ exactly when $T$ is not invertible with bounded inverse.