On Minimal polynomial and Invariant subspace

Question

Given $T:V \rightarrow V$, $V$ is a vector space over field $\mathbb{R}$ and $m_T = (x^2-2x+2)(x-3)^2$. Show that there exists an invariant subspace with dimension $2$. I first thought that since $3$ is an eigenvalue and $x-3$ appears in a power of $2$, then $V_3$ might have dimension $2$ and we're done but thinking more about it I now think this is not necessarily the case. I also tried using: $$V = \ker(T^2-2T+2I) + \ker ((T-3)^2)$$
but this gives a dimension of at least $2$ and not exactly $2$

Answer 1

here's my proof, correct? If $dimV_3>1$ its easy. otherwise the dimension is 1. assume for contradcition $dimker((T−3I))^2=1$ to get (define $S = T^2-2T+2I$): $dimKer(T^2−2T+2I)=dimKer(S)=dimV−1=n−1$. take a basis for that space $B=v_1,...,v_{n−1}$. and take an eigenvector with eigenvalue 3, $v_n$. now $v_1,...,v_n$ is a basis for V, because if $a_1v_1+...+a_nv_n=0$ that implies $a_n=0$ by taking S of both sides. therefore all the ${a_i}$ are $0$ because $B$ is a basis. Now $f(x)=(x^2−2x+2)(x−3)$ satisfies $f(T)=0$ because each of the basis vectors is in the kernel of at least one of $T−3I,T^2−2T+2I$.

now that's a contradiction because $deg(f)=2<3=degm_T$. so we get $dim(ker(T−3)^2)>1$. since $dimV_3 = 1$ there exists a vector $v∈ker(T−3)^2$ and $v∉V_3$. This vector is not an eigenvector and therefore $T(v)$ is not a multiple of $v$, but since $(T−3I)^2(v)=0$ we get $T^2(v)=6T(v)−9v$ and therefore, $W=sp(v,T(v))$ is T-Invariant. Is this proof correct? if you had another proof, I'd like to read it too. thanks!

Answer 2

It is not difficult to show the much more general following result. Let $n\geq 2$ and $A\in M_n(\mathbb{R})$.

$\textbf{Proposition}$ $A$ admits an invariant real space of dimension $2$.

$\textbf{Proof}$. Case 1. $A$ admits at least $2$ distinct real eigenvalues. It is easy...

Case 2. $A$ admits one eigenvalue $s$ of multiplicity $n$; necessarily $s$ is real and we may assume that $s=0$, that is, $A$ is nilpotent. If $\ker(A)$ has dimension $\geq 2$, then we are done; if $\ker(A)$ has dimension $1$, then $\ker(A^2)$ has dimension $2$ and we are done.

Case 3. $A$ admits (at least) $2$ non-real conjugate eigenvalues $s,\overline{s}$. If $u$ is an eigenvector associated to $s$, then $\overline{u}$ is an eigenvector associated to $\overline{s}$. Then consider $span((u+\overline{u}),(u-\overline{u})/i))$.

Answer 3

Probably easiest here is to use the irreducible factor $x^2-2x+2$ of the minimal polynomial. The kernel of $T^2-2T+2I$ has nonzero dimension (otherwise that factor would be absent from the minimal polynomial), and for any nonzero vector $v$ of that kernel the subspace spanned by $v$ and $T(v)$ is invariant (since $T^2(v)=2T(v)-2v$) and of dimension$~2$ (since $v$ is not annihilated by any polynomial of degree${}<2$ in $T$, which polynomial would have to divide $x^2-2x+2$).

I should note that irreducibility of $x^2-2x+2$, while convenient, is not essential to the argument: for any monic polynomial divisor$~D$ of the minimal polynomial there are vectors that are annihilated by $D[T]$ but not by any lower degree monic polynomial of $T$, and under the action of $T$ such a vector spans an invariant subspace of dimension $\deg(D)$. In particular you can apply this in the example for $D=(x-3)^2$ too. And of course, since over $\Bbb R$ all irreducible polynomials have degree $1$ or $2$, any minimal polynomial of degree at least$~2$ has a divisor of degree exactly$~2$, from which it follows that for any linear operator $T$ on a finite dimensional real vector space of dimension at least$~2$ there exists a $2$-dimensional $T$ invariant subspace, as mentioned in the answer by loup blanc.