$V$ 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.
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.
$\textbf{Edit}.$ We can generalize the above result as follows:
$\textbf{Proposition}$.Let $K$ be a field and $A\in M_n(K)$. 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 $\leq 7$.
The converse, in fact, also holds. We shall extend this to the general case on linear maps (Linear Algebra Done Right 4th Edition, 5A.39, p.142)
Suppose $V$ is a finite-dimensional (vector space over $\mathbb{F}\! =\!\mathbb{R}$ or $\mathbb{C} $), and $T\!\in\!\mathcal{L}(V).$ Prove that $T$ has an eigenvalue if and only if there exists a subspace of $V$ of dimension $\text{dim}\, V\! -\! 1$ that is invariant under $T.$
$Proof:$
Let $n\! =\!\text{dim}\, V$ ; $\, N_T\! =\! \text{null }T$ ; $\, R_T\! =\!\text{range }T$ ; $\, m\!=\!\text{dim}\, N_T.$
$ $
If $\,\mathcal{U}\,$ is a subspace of $V$ invariant under $T,$ the quotient map $T/\mathcal{U},$ $$T/\mathcal{U}(v+\mathcal{U}) = Tv+\mathcal{U},$$ is then well-defined and an operator on the quotient space $V/\mathcal{U}.$
$ $
A space is invariant under $T$ if and only if it is invariant under $p(T)$ for every polynomial. In particular, being invariant under $T$ is equivalent to being invariant under $T\! -\!\lambda I$ for some/any $\lambda\!\in\!\mathbb{F}.$
$ $
(⇒) Without loss of generality, let $\lambda\! =\! 0$ be an eigenvalue of T (If $\lambda\!\not=\! 0,$ let $S\! =\! T\! -\! \lambda I$. By the paragraph above, it suffices to find a $(\text{dim}\, V\! -\! 1)$-dimensional subspace of $V$ invariant under $S $).
$ $
If $N_T = N_{T^2},$
$\iff N_T \cap R_T\! =\! \{0\}\iff V\! =\! N_T\oplus R_T,$
where the first equivalence holds because $T(N_{T^2})\! =\! N_T\cap R_T$ and $N_T\!\subset\! N_{T^2},$ while the last one holds because $\text{dim}\, V = \text{dim}\, N_T + \text{dim}\, R_T = \text{dim}(N_T\oplus R_T).$
$ $
Let $\{y_1,...,y_k\}$ be a basis of $R_T,$ and $\{x_1,...,x_m\}$ be that of $N_T.$
$\implies\!\! \{x_1,...,x_m,y_1,...,y_k\}$ is a basis of $V$ (due to the direct sum decomposition above).
$ $
Consider any $E\subset \{x_1,...,x_m\}$ with $|E|\! =\! n\! -\! k\! -\! 1.$ Then $\text{span}(E\cup \{y_1,...,y_k\})$ is of dimension $\text{dim}\, V\! -\! 1$ and invariant under $T.$
$ $
Else if $V=N_T,$ then the span of any list of $\text{dim}\, V\! -\! 1$ vectors subset of a basis of $V$ suffices.
$ $
Otherwise (that is if $N_T\!\not=\! N_{T^2}$ and $V\!\not=\! N_T $), we shall pave the way by induction.
The forward implication trivially holds for $\text{dim}\, V\!\in\!\{1,2\}.$
$ $
Assume that $\text{dim}\, V > 2.$
Suppose that the forward implication holds for all nonzero vector spaces of dimension less than $\text{dim}\, V.$
$ $
\begin{align*} N_T\!\not=\! N_{T^2}\!\iff &N_T \subsetneq N_{T^2}\\ \iff &N_{T^2}\!\!\setminus\! N_T\text{ contains a nonzero vector in } V,\end{align*} where the last equivalence holds because $0\not\in N_{T^2}\!\!\setminus\! N_T\not=\emptyset.$
$ $
Let $v\!\in\! N_{T^2}\!\!\setminus\! N_T$ (so $v\!\not=\! 0$ by the previous reasoning). \begin{align*} \implies &Tv∈N(T)\\ \implies & T/N_T(v+N_T)=Tv+N_T=N_T\\ &\text{with } v+N_T\not=N_T\: (\text{since }v\!\not\in\! N_T)\\ \implies &T/N_T \text{ has an eigenvalue}\end{align*}
($N_T$ is invariant under $T$, so the consideration of $T/N_T$ is unambiguous by the preface).
$ $
$$V\not= N_T\not=\{0\}\!\implies\! 0<\text{dim}(V/N_T)=\text{dim}\, V - \text{dim}\, N_T < \text{dim}\, V.$$ By induction hypothesis, there is an $(n\! -\! m\! -\! 1)$-dimensional subspace of $V/N_T$ invariant under $T/N_T.$
Let $\{u_1+N_T,...,u_{n-m-1}+N_T\}$ be a basis of the said subspace.
Unraveling the meaning behind this, \begin{align*} &\forall\! _ {j\in\{1,...,n-m-1\}}\! : \, Tu_j+N_T = T/N_T(u_j+N_T)\in \text{span}(u_1+N_T,...,u_{n-m-1}+N_T) \\ \iff &\forall\! _ {j\in\{1,...,n-m-1\}}\! : \, Tu_j \in \text{span}(u_1,...,u_{n-m-1})+N_T=W,\end{align*}
which, along with the fact that $T$ sends $N_T$ to $\{0\} ,$ shows that $W$ is invariant under $T.$
$ $
The linear-independence of $\{u_1+N_T,...,u_{n-m-1}+N_T\}$ in $V/N_T$ implies that of $\{u_1,...,u_{n-m-1}\}$ in $V$ and that each $u_j\!\not\in\! N_T.$ \begin{align*}\implies &\{u_1,...,u_{n-m-1}\}\cup B\text{ is linearly-independent in $V$ for every basis }B\text{ of }N_T \\ \implies &\text{dim}\, W = (n-m-1)+\text{dim}\, N_T = \text{dim}\, V - 1\end{align*}
(to validate the linear-independency of $\{u_1,...,u_{n-m-1}\}$ in $V,$ note that if $A$ is any linear map from one vector space to another, the linear independence of $\{Av_1,...,Av_j\}$ in the codomain implies that of $\{v_1,...,v_j\}$ in the domain. Consider $\pi(v)=v+N_T $).
In all cases, we can find such subspace of $V.$
$ $
(⇐) Let $U$ be the said subspace. Extending $U$ to $V,$ there is a $v\!\in\! V\!\setminus\! U$ such that $V=U\oplus\text{span}(v).$
\begin{align*} \implies &Tv=\lambda v+u \text{ for some }\lambda\!\in\!\mathbb{F}, \, u\!\in\! U \\ \iff &(T-\lambda I)v \in U\\ \iff &(T-\lambda I)(\text{span}(v)) \subset U, \end{align*}
and so by hypothesis,
\begin{align*} &U \text{ invariant under } T \\ \iff &U\text{ invariant under }T-\lambda I\\ \iff &(T-\lambda I)(U) \subset U\\ \implies &R_{T-\lambda I}=(T\! -\!\lambda I)(U)+(T\! -\! \lambda I)(\text{span}(v)) \subset U\subsetneq V\\ \implies &T-\lambda I\text{ is not surjective}\\ \iff &T-\lambda I\text{ is not injective}\\ \iff &\lambda\text{ is an eigenvalue of }T. \end{align*} $$\tag*{$\blacksquare$}$$
