Existence of $n-1$ dimensional invariant subspace of $V$ over $\mathbb{R}$ given characteristic polynomial has a real root. $V$ is a finite dimensional vector space over $\mathbb{R}$ with $\dim V \ge 1$ and $\phi \in L(V, V)$ is an endomorphism. Its characteristic polynomial $w_{\phi}(\lambda)$ has a real root. Prove the existence of an $n-1$ dimensional invariant subspace for $V$.
I tried to deduce something from Jordan Canonical Form but with no effect.
 A: There is $\lambda\in\mathbb{R},v\in\mathbb{R}^n\setminus\{0\}$ s.t. $A^Tv=\lambda v$. 
Then $v^{\perp}=\{u\in\mathbb{R}^n;v^Tu=0\}$ is a $A$-invariant subspace of dimension $n-1$.
$\textbf{Proof}.$ Indeed, if $u\in v^{\perp}$, then $v^T(Au)=(v^TA)u=\lambda v^Tu=0$.
Remark. In fact, $A$ admits invariant subspaces of any dimension.
EDIT. We can generalize the above result as follows. Let $K$ be a field and $A\in M_n(K)$.
$\textbf{Proposition}$. If $\lambda\in K$ is an eigenvalue of $A$ of algebraic multiplicity $k$, then $A$ admits invariant $K$-subspaces of dimensions $1,\cdots,k$ and $n-k,\cdots,n-1$.
$\textbf{Proof}$. i) $\chi_A(x)=(x-\lambda)^kf(x)$ where $f(x)\in K[x]$. Up to a change of basis in $K^n$, we may assume that $A=diag(U_k,B_{n-k})$ where $U-\lambda I_k$ is triangular nilpotent and $\chi_B=f$. Thus $span(e_1),span(e_1,e_2),\cdots,span(e_1,\cdots,e_k)$ are $A$-invariant.
ii) According to i), for every $p\leq k$, $A^T$ admits an invariant  vector space of dimension $p$, say $V$. Then $V^{\perp}=\{u\in K^n;\text{ for every }v\in V,v^Tu=0\}$ is a $A$-invariant $K$-subspace of dimension $n-p$ (here, $v^Tu$ is not a scalar product). Indeed, if $u\in V^{\perp}$, then $A^Tv\in V$ and $v^TAu=0$.  $\square$
Example. $K=\mathbb{Z}/3\mathbb{Z}$, $\chi_A(x)=(x^3-1)(x^4+x^3+x^2+1)$. $A$ admits invariant $K$-subspaces of any dimension $\leq 7$.
