$V$ 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.

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    $\begingroup$ The image of $(\phi - \lambda I)$ is an invariant subspace, but the dimension of this subspace is $n - d$ where $d$ is the geometric multiplicity of the eigenvalue $\lambda$. $\endgroup$ – Ben Grossmann Jun 18 '19 at 17:14
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    $\begingroup$ Thanks, I get it. So we can take $(\phi - \lambda I)$, then $\ker(\phi - \lambda I) \neq \{0\}$ so $\dim im(\phi - \lambda I) \le n - 1$. Now any $n - 1$ dimensional subspace $V'$ such that $im(\phi - \lambda I) \subset V'$ is invariant under $(\phi - \lambda I)$ so is invariant under $\phi$ $\endgroup$ – user4201961 Jun 18 '19 at 17:37

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 $\leq 7$.

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