complex eigenvalues and invariant spaces

Question

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.

Answer 1

No, the space spanned by $v_1$ and $v_2$ is not the space you are looking for: $v_1$ and $v_2$ have complex entries in fact.
So we assume that $v_2=\bar v_1,$ and consider $w_1=v_1+\bar v_1$ and $w_2=(v_1-\bar v_1)i\in\mathbb R.$
Then we see that $$E(w_1)=aw_1+bi(v_1-\bar v_1)=aw_1+bw_2$$ $$E(w_2)=aw_2-b(v_1+\bar v_1)=-bw_1+aw_2.$$
Thus the space spanned by $w_1$ and $w_2$ is invariant under $E.$
Hope this helps.

Answer 2

As you indicated the result is easy if there are any real eigenvalues, so I'll focus on the contrary case.

Note that your complex eigenvectors don't live in $\Bbb R^k$ at all, but in $\Bbb C^k$. In order to get back to $\Bbb R^k$, you may view it as a real subspace of $\Bbb C^k$ (vectors with all coordinates real) and try to construct from the eigenspaces in$~\Bbb C^k$ an invariant subspace in $\Bbb R^k$. It turns out that the complex span of a single eigenvector intersects the real subspace $\Bbb R^k$ trivially, but the complex span of an eigenvector together with its component-wise complex conjugate, which is an eigenvector for the conjugate eigenvalue, does intersect $\Bbb R^k$ in a real $2$-dimensional subspace, invariant under the real linear operator. In fact the component-wise real and imaginary "parts" of a complex eigenvector $v$ (which van be obtained as linear combination of $v$ and its conjugate) span such an invariant subspace. The answer by @awllower details this approach.

There is another approach that avoids complexifying the vector space to begin with. The minimal polynomial $P$ of the linear operator $E$ has real entries (so does its characteristic polynomial, which could be used in this argument as well, invoking the Cayley-Hamilton theorem). It can therefore be decomposed as product $P=P_1\ldots P_m$ of monic irreducible polynomials over the real numbers, where each factor $P_i$ has degree $1$ or $2$ (due to the FTA). Now by definition of the minimal polynomial $P[E]=0$, which becomes $$ P_1[E]\circ P_2[E]\circ\cdots\circ P_m[E]=0. $$ Clearly at least one of the $P_i[E]$ fails to be injective (in fact they all do). Let $0\neq v\in\ker(P_i[E])$. If $\deg(P_i)=1$ then $P_i=X+c$ for some $c\in\Bbb R$ and $v$ is an eigenvector for$~{-}c$ (the easy case). Otherwise $P_i=X^2+bX+c$ for some $b,c\in\Bbb R$, so $E^2(v)+bE(v)+cv=0$ or equivalently $E^2(v)=-bE(v)-cv$ which shows that the span of $v$ and $E(v)$ is an invariant subspace for $E$.

Answer 3

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?