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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 .

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    $\begingroup$ Consider Jordan blocks. Alternatively, if you have two $1$-dimensional subspaces, you can take their sum. $\endgroup$ Oct 26, 2014 at 13:12
  • $\begingroup$ @NajibIdrissi: But does $T$ necessarily have two $1$ dimensional invariant subspace ? $\endgroup$
    – Souvik Dey
    Oct 26, 2014 at 13:16
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    $\begingroup$ Try to assume it only has one and see what happens. $\endgroup$ Oct 26, 2014 at 13:18
  • $\begingroup$ @NajibIdrissi: Could you please elaborate ? $\endgroup$
    – Souvik Dey
    Oct 26, 2014 at 13:24

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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.

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  • $\begingroup$ In the $s \ge 2$ case, how do you know that $\mathrm{span}(e1, e2)$ is two-dimensional? (E.g., what if $\lambda_1 = \lambda_2$?) $\endgroup$ Nov 2, 2014 at 9:54
  • $\begingroup$ @MarkDickinson Corrected, thanks. $\endgroup$ Nov 2, 2014 at 10:06
  • $\begingroup$ @EwanDelanoy Can you explain why ${\sf Ker}((f-\lambda{\textsf{id}})^p)=W_1$ for every $p$? I understood that by induction we need to prove that $W_p=W_1$. The direction $W_1 \subset W_p$ is easy to prove. How about $W_p \subset W_1$? $\endgroup$
    – user
    Apr 5, 2015 at 2:46
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    $\begingroup$ @user please see my little update $\endgroup$ Apr 5, 2015 at 5:35
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    $\begingroup$ @user ${\mathbb R}^n={\textsf{Ker}}((T-\lambda \textsf{id})^{s_1})$ by the Cayley-Hamilton theorem $\endgroup$ Apr 21, 2015 at 8:37

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