Complexification and spectrum of a real operator

Assume the Axiom of Choice (The Axiom of Choice), inherited by the spectral-radius definition. Let $X$ be a nonzero real Banach space and let $T:X\to X$ be a bounded real-linear operator (A bounded linear operator between normed spaces). Let $X_{\mathbb{C}}=X\times X$ be the canonical complexification with the rotation-supremum norm $\rho$ and let $T_{\mathbb{C}}(x,y):=(Tx,Ty)$ be the complex-linear extension, both from Canonical Banach complexification of a real Banach space; thus $T_{\mathbb{C}}\in \mathcal{B}(X_{\mathbb{C}})$ with $\mathrm{\parallel }{T}_{\mathbb{C}}\mathrm{\parallel }=\mathrm{\parallel }T\mathrm{\parallel }$. The spectrum, resolvent set, resolvent and spectral radius of the real operator $T$ are those of $T_{\mathbb{C}}$ computed in the unital Banach algebra $\mathcal{B}(X_{\mathbb{C}})$:

$\sigma (T):={\sigma }_{\mathcal{B}(X_{\mathbb{C}})}(T_{\mathbb{C}}),\rho (T):=\mathbb{C}\setminus \sigma (T),R(z,T):=(z1-T_{\mathbb{C}}{)}^{-1}(z\in \rho (T)),r(T):=r(T_{\mathbb{C}}),$

with the conventions of Spectrum and resolvent set in a Banach algebra and Spectral radius.

The operator algebra is complete by If (Y) is Banach then (\mathcal B(X,Y)) is Banach. Composition is bilinear and associative, and $\mathrm{\parallel }STu\mathrm{\parallel }\le \mathrm{\parallel }S\mathrm{\parallel }\mathrm{\parallel }T\mathrm{\parallel }\mathrm{\parallel }u\mathrm{\parallel }$ gives submultiplicativity by The operator norm as the least bound and as the unit-sphere or unit-ball supremum. Its identity has norm one: it is bounded by one, and a nonzero vector, normalized to norm one, gives equality. Thus it is nonzero and satisfies Unital Banach algebra. The same argument applies to each comparison model below, since its embedded real copy is nonzero.

Well-definedness (independence of the complexification model). Let $Z$ be another compatible complexification of $X$ in the sense of claim 3 of Canonical Banach complexification of a real Banach space: a complex Banach space with a real-linear isometric embedding $j_Z:X\to Z$ such that $Z=j_Z(X)\oplus i{j}_Z(X)$, an isometric conjugation ${\sigma }_Z$, and let $T_Z(j_Z(x)+i{j}_Z(y)):=j_Z(Tx)+i{j}_Z(Ty)$ be the corresponding extension of $T$. By that lemma the map $\mathrm{\Phi }:X_{\mathbb{C}}\to Z$, $\mathrm{\Phi }(x,y)=j_Z(x)+i{j}_Z(y)$, is a bounded complex-linear bijection with bounded inverse ${\mathrm{\Phi }}^{-1}$, and $\mathrm{\Phi }{T}_{\mathbb{C}}{\mathrm{\Phi }}^{-1}=T_Z$. Hence for every $z\in \mathbb{C}$

$z1-T_Z=\mathrm{\Phi }(z1-T_{\mathbb{C}}){\mathrm{\Phi }}^{-1},$

so $z1-T_Z$ is invertible in $\mathcal{B}(Z)$ exactly when $z1-T_{\mathbb{C}}$ is invertible in $\mathcal{B}(X_{\mathbb{C}})$, with $(z1-T_Z{)}^{-1}=\mathrm{\Phi }(z1-T_{\mathbb{C}}{)}^{-1}{\mathrm{\Phi }}^{-1}$. Taking spectra,

${\sigma }_{\mathcal{B}(Z)}(T_Z)={\sigma }_{\mathcal{B}(X_{\mathbb{C}})}(T_{\mathbb{C}}),r(T_Z)=r(T_{\mathbb{C}}),$

because the bijection $z\mapsto z$ matches the two spectral sets and preserves moduli. So the spectrum and the spectral radius of a real operator do not depend on which compatible complexification computes them, and all of them are computed below in the canonical model.

• Why not "real $\lambda$ with $\lambda I-T$ not invertible". Restricting the discussion to real scalars would discard the genuinely complex part of the spectrum. For example the quarter-turn $J(u,v)=(-v,u)$ on Euclidean ${\mathbb{R}}^2$ has complexified spectrum exactly $\{i,-i\}$: $J^2=-I$, so for $z^2+1\mathrm{\ne }0$ the inverse of $zI-J$ is $(zI+J)\mathrm{/}(z^2+1)$; at $z=i,-i$ the respective nonzero complex vectors $(1,-i)$ and $(1,i)$ are in the kernel. Its real-scalar resolvent is all of $\mathbb{R}$, whereas its complex spectrum is nonempty. More precisely the real-scalar noninvertibility set equals $\sigma (T)\cap \mathbb{R}$. Indeed, a bounded real inverse extends componentwise to a bounded complex inverse. Conversely, for real $z$ the operator $zI-T_{\mathbb{C}}$ commutes with the canonical conjugation $C(x,y)=(x,-y)$, which is isometric by replacing $\theta$ with $-\theta$ in the norm formula. Its bounded inverse therefore also commutes with $C$ and restricts to a bounded inverse on its fixed real copy $X\times \{0\}$. This gives both directions without conflating real and complex spectra.

• The operator is bounded by hypothesis. The same-norm extension statement of the complexification lemma is used only for bounded real-linear $T$; it is what makes $T_{\mathbb{C}}$ an element of the Banach algebra $\mathcal{B}(X_{\mathbb{C}})$, to which the spectral theory of this page applies. For unbounded real operators no spectrum in this sense is defined here.

• $z\mapsto R(z,T)$ is a holomorphic $\mathcal{B}(Z)$-valued map on $\rho (T)$ by Resolvent is Banach-valued holomorphic, applied in whichever model is used; the comparison isomorphism above conjugates one resolvent map into the other. In particular the resolvent of a real operator is well defined at a point $z\in \mathbb{C}$ exactly when $z\mathrm{\notin }\sigma (T)$.