spectrum of an operator restricted to an invariant subspace Let $X$ be an infinite-dimensional real Banach space and $T\in\mathcal{L}(X)$ a continuous linear operator acting on $X$.  Suppose $W$ is a finite-codimensional $T$-invariant closed subspace of $X$, and write $S=T|_W\in\mathcal{L}(W)$ for the restriction of $T$ to $W$.
Denote by $\sigma(T)$ the spectrum of the complexification $T_\mathbb{C}\in\mathcal{L}(X_\mathbb{C})$, and similarly for $S$.  In other words, $\lambda\in\sigma(T)$ if and only if $(\lambda-T_\mathbb{C})$ is not invertible in the complex operator algebra $\mathcal{L}(X_\mathbb{C})$.
Conjecture.  If $\sigma(T)\subseteq\mathbb{R}$ then $\sigma(S)\subseteq\mathbb{R}$.
Note.  For my purposes we can assume $X$ is reflexive and that $W=\overline{p(T)X}$ for some polynomial $p(t)\in\mathbb{R}[t]$ (with real coefficients).
I don't want to make any additional assumptions about $W$ or $S$.  However, if necessary we could make additional assumptions on $X$ and/or $T$, for instance that $X$ is a Hilbert space and that $T$ is selfadjoint.  Of course, the fewer assumptions the better.
 A: $W_C = \{u + iv \in X_C \mid u, v \in W\}$ is invariant under $T_C$, so we may as well forget about the real Banach spaces and consider an operator $T$ on a complex Banach space $X$ with a finite-codimensional invariant subspace $W$, such that $\sigma(T) \subseteq \mathbb R$.  The claim is that $\sigma(T|_W) \subseteq \mathbb R$.
If $\lambda \notin \sigma(T)$, $T - \lambda I$ is invertible as an operator on $X$.  Let $W$ have codimension $k$ in $X$.  Thus $X$ is spanned by $W$ and $k$ vectors $v_1, \ldots, v_k$, with $W \cap \text{span}(v_1, \ldots, v_k) = \{0\}$  Since $T-\lambda I$ is surjective, $X$ is also spanned by $(T-\lambda I)W$ and the $k$ vectors $(T-\lambda I)v_1, \ldots, (T-\lambda I)v_k$, with $(T-\lambda I) W \subseteq W$. Thus $(T-\lambda I) W$ still has codimension $k$, and so $(T-\lambda I) W = W$.  Thus $(T-\lambda I)^{-1}$ (as an operator on $X$) is also an inverse for $T -\lambda I$ on $W$.  This says that $\sigma(T|_W) \subseteq \sigma(T) \subseteq \mathbb R$.
EDIT: In general, without the finite-codimension assumption $\sigma(T|_W)$ might be larger than $\sigma(T)$.  However, it is always true that if $\lambda \notin \sigma(T) \cup \sigma(T|_W)$, $(T|_W - \lambda I|_W)^{-1} = (T-\lambda I)^{-1}|_W$.  Now if $\lambda$ is in the boundary of $\sigma(T|_W)$, we can take a sequence $\lambda_n \to \lambda$ with $\lambda_n \notin \sigma(T|_W)$, and we must have $\| (T|_W - \lambda_n I|_W)^{-1}\| \to \infty$.  If $\sigma(T) \subseteq \mathbb R$, we can take $\lambda_n \notin \mathbb R$, so these are also not in $\sigma(T)$, and thus 
$\|(T - \lambda_n I)^{-1}\| \ge \| (T - \lambda_n I)^{-1}|_W\|  \to \infty$, which implies $\lambda \in \sigma(T)$.  Thus the boundary of $\sigma(T|_W)$ is contained in $\mathbb R$, and this implies $\sigma(T|_W) \subset \mathbb R$.  So the finite-codimension assumption is not needed for this.
