Existence of invariant subspaces of $\mathbb C^n$ and $\mathbb R^n$ Exercise $6.15$ from Dym's Linear Algebra in Action:

Let $A \in \mathbb R^{n\times n}$ and suppose that $n \geq 2$. Show that: 
$(1)$ There exists a one-dimensional subspace U of $\mathbb C^n$ that is invariant 
  under $A$. 
$(2)$ There exists a subspace $V$ of $\mathbb R^n$ of dimension less than or equal to 
  two that is invariant under $A$.

First of all: does he mean  $\dim_\mathbb C U$ and $\dim_\mathbb R V$?
For $(1)$: the Jordan form surely exists, possibly with $J, S \in \mathbb C^{n\times n}$. But what do we know about one-dimensional subspaces if, say, no eigenvalue has one-dimensional eigenspace?
For $(2)$: the exercise comes after a short section on Jordan decompositions for real matrices, so I guess one can read it from the real Jordan form. What would be the good argument?
 A: I'm pretty sure he means $\dim_{\Bbb C}U$ and $\dim_{\Bbb R}V$, and I will assume so in this answer, for reasons which will become more clear in what follows.
For (1), consider $A:\Bbb C^n \to \Bbb C^n$, i.e., the matrix $A$ as it acts on $\Bbb C^n$.  Since $\Bbb C$ is algebraically closed, all $n$ roots of the characteristic polynomial $p_A(\lambda) = \det(A - \lambda I)$ exist in $\Bbb C$, and if we pick any specific eigenvalue $\lambda_s \in \Bbb C$, then there must exist a nonzero vector $\vec v \in \Bbb C^n$ with ($A - \lambda_s I)\vec v_s = 0$, or $A \vec v = \lambda_s \vec v$, since $\det(A - \lambda_s I) = 0$.  Then the one-dimensional subspace $\Bbb C \vec v = \{\mu v \mid \mu \in \Bbb C \}$ is invariant under $A$:  $A(\mu \vec v) = \mu A\vec v = \mu (\lambda_s \vec v) = (\mu \lambda_s)\vec v \in \Bbb C \vec v$ for any $\mu \in \Bbb C$.  There is at least one such subspace for every eigenvalue of $A$.
Note that the above conclusion may not hold if we mean $\dim_{\Bbb R}U$, since the (real) one-dimensional subspace $\Bbb R \vec v$ will not generally be invariant under $A$ if the eigenvalue $\mu$ corresponding to $\vec v$ is complex; with $r \in \Bbb R$, $A(r \vec v) = r\mu \vec v$: but in general there is no real $s$ with $s \vec v = r \mu \vec v$; if there were, we would have $(s - r \mu) \vec v = 0$ forcing $s = r\mu$, a contradiction.  This leads me to give credence to the assumption that $\dim_{\Bbb C}U$ is intended.
As for (2), considering $A:\Bbb R^n \to \Bbb R^n$, we see that if $\mu \in \Bbb R$ is an eigenvector of $A$ with eigenvector $\vec v \in \Bbb R^n$, then $\Bbb R \vec v$ is $A$-invariant, by an argument similar to that given above, and $\Bbb R \vec v$ is clearly one dimensional over $\Bbb R$.  If $\mu \in \Bbb C$ is an eigenvalue of $A$, we must have $A \vec v = \mu \vec v$ for some $\vec v \in \Bbb C^n$; setting $\vec v = \vec r + i \vec s$, where $\vec r, \vec s \in \Bbb R^n$, we see that the equation $A\vec v = \mu \vec v$ becomes
$A \vec r = \Re(\mu) \vec r - \Im(\mu) \vec s, \; A\vec s = \Re(\mu) \vec s + \Im(\mu) \vec r, \tag{1}$
which clearly implies the subspace $\text{span} \{ \vec r, \vec s \} = \{ a \vec r + b \vec s \mid a, b, \in \Bbb R \}$ is $A$-invariant; being spanned by $\vec r$, $\vec s$, it is clearly of real dimension $2$.  And this fact enforces my belief that $\dim_{\Bbb R}V$ is meant; in any event, $V$ is a real vector space, so $\dim_{\Bbb C}V$ is not defined without specifying the action of $\Bbb C$ on $V$ in a manner consistent with the vector space axioms.  QED.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
