complex eigenvalues and invariant spaces I am currently reading Guillemin and Pollack's Differential Topology, and the following claim is made without proof: 
Given a linear isomorphism $E: \mathbb{R}^k \to \mathbb{R}^k$, with $k>2$ and such that $E$ can be represented by a matrix with real entries, $E$ has a one- or two-dimensional invariant space.
I understand that the Fundamental Theorem of Algebra implies that $E$ has at least one real or complex eigenvalue; if it is real, then $E$ clearly has a one-dimensional fixed space. If it is complex, however, I don't see how there needs to be a two-dimensional invariant space.
If $E$ has complex eigenvalue $a+bi$, then $a-bi$ must also be an eigenvalue (as $E$ contains real entries). These eigenvalues correspond to eigenvectors $v_1$ and $v_2$. I assume that the subspace spanned by $v_1$ and $v_2$ is the desired invariant space, but can't figure out how to prove it. Any help would be most appreciated.

I think here is another way. A has eigenvalue $\lambda$ and eigenvector $\textbf{v}\in \mathbb{C}^n$. We can say $\lambda = a+bi$ and $v_j=x_j+iy_j \text{ for }j=1,2,...,n\Rightarrow \textbf{v}={\textbf{x}}+i\textbf{y}  \text{ where }\textbf{x,y}\in \mathbb{R}^n$ . So now we have: $$(1)\quad A\textbf{v}=\lambda \textbf{v}=\lambda (\textbf{x}+\textbf{y}i)=(a+bi)(\textbf{x}+\textbf{y}i)=(a\textbf{x}-b\textbf{y})+i(b\textbf{x}+a\textbf{y})\\$$
But now you also know that $$(2) \quad A\textbf{v}=A(\textbf{x}+\textbf{y}i)=A\textbf{x}+A\textbf{y}i.\\$$
So I can set (1) and (2) equal:
$$A\textbf{x}+A\textbf{y}i=(a\textbf{x}-b\textbf{y})+i(b\textbf{x}+a\textbf{y})$$
But now we can set the real and imaginary parts equal. We have:
$$A\textbf{x}=(a\textbf{x}-b\textbf{y}) \text{ and } A\textbf{y}=(b\textbf{x}+a\textbf{y}), \text{ where }A\textbf{x}, A\textbf{y} \in \mathbb{R}^n$$
So the invariant subspace is the plane spanned by $\textbf{x}$ and $\textbf{y}$ and is two dimensional.
Could you help me understand the connection between my x and y and your $w_1$ and $w_2$. I think my $2x=w_1$ and $2y=w_2$ so the span is the same, but then when we multiply by A the stuff is different? How?
