For $n \ge 2$ , does every linear operator on $\mathbb R^n$ has an invariant subspace of dimension $2$ ? Is it true that for $n \ge 2$ , every linear operator $T$ on $\mathbb R^n$ has an invariant subspace of dimension $2$ ? I know that $T$ always either have a $1$ or $2$ dimensional invariant subspace ; but cannot determine whether it will always have a $2$ dimensional one . Please   help . 

To elaborate on Najib Idrissi’s comment :
The answer is YES. Indeed, the characteristic polynomial $\chi_T$ of $T$ can be factorized over $\mathbb R$ as
$$
\chi_T(x)=\prod_{k=1}^{s} (x-\lambda_k)^{s_k} \prod_{j=1}^t ((x-a_j)^2+b_j^2)
$$
where $\sum_{k=1}^{s} s_k+2t=n$, all $\lambda_k,a_j,b_j$ real, $\lambda_{k}\neq \lambda_{k'}$ when $k\neq k'$, and $b_j\neq 0$ for each $j$.
If $s\geq 2$, then as Najib noted taking an eignevector $e_1$ for $\lambda_1$
and an eigenvector $e_2$ for $\lambda_2$, considering the two-dimensional
space ${\textsf{span}}(e_1,e_2)$ we are done.
Similarly, if $t\geq 1$, $a_j+b_ji$ is a complex eigenvalue for $T$, so there
is a complex eigenvector $u+iv$ where $u$ and $v$ have real coordinates.
The eigenidentity $T(u+iv)=(a_j+ib_j)(u+iv)$ yields $Tu=a_ju-b_jv$ and
$Tv=b_ju+a_jv$. It follows that $u$ and $v$ are linearly independent and that
the subspace ${\textsf{span}}(u,v)$ is invariant by $T$.
So if $T$ has no two-dimensional invariant subspace, we must have
$s<2,t<1$, and hence
$$
\chi_T(x)=(x-\lambda)^{s_1}
$$
for some $\lambda\in {\mathbb R}$. 
If $W_1={\sf Ker}(T-\lambda{\textsf{id}})$ has dimension at least $2$, then
any two-dimensional subspace of $W_1$ will satisfy us. So we can assume
that $W_1$ has dimension $1$, $W_1={\textsf{span}}(e_1)$ for some vector
$e_1$. Consider now $W_2={\sf Ker}((T-\lambda{\textsf{id}})^2)$. If $W_2=W_1$, then $W_{p+1}=W_p$ for every $p \geq 1$ (indeed, if $w\in W_{p+1}$ then $w'=(T-\lambda{\textsf{id}})^{p-1}(w)\in W_2$, so $w'\in W_1$ and hence $w\in W_p$) 
so ${\sf Ker}((T-\lambda{\textsf{id}})^p)=W_1$ for every $p$,  and hence ${\mathbb R}^n={\sf Ker}((T-\lambda{\textsf{id}})^{s_1})=W_1$ which 
is impossible if $n>1$. So $W_1\neq W_2$. Since $W_1$ is a subspace of
$W_2$, we see that there is a vector $e_2\in W_2\setminus W_1$. We have
$(T-\lambda {\sf id})e_2\in W_1$ by definition of $W_1$ and $W_2$, so 
${\textsf{span}}(e_1,e_2)$ is invariant by $T$. This concludes the proof.
