# 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.

• 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 $\lambda$. – Ben Grossmann Jun 18 '19 at 17:14
• 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 $\phi$ – 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 $$\leq 7$$.