# On Minimal polynomial and Invariant subspace

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 $$2$$

• The exponent of the irreducible in the minimal polynomial does not tell you the dimension of the eigenspace. For instance, if $V=\mathbb{R}^{10}$ and $T=I$, then the minimal polynomial is $m_T=(x-1)$, but $V_1$ has dimension $10$, not $1$. – Arturo Magidin Apr 5 '19 at 22:18
• Consider $\mathbf{R}((T-3I)^2)$. Prove that it is not zero, and contains a nonzero vector $\mathbf{v}$ such that $T(\mathbf{v})$ is not a multiple of $\mathbf{v}$, but such that $T^2(\mathbf{v})\in\mathrm{span}(\mathbf{v},T(\mathbf{v}))$. – Arturo Magidin Apr 5 '19 at 22:27
• @ArturoMagidin here's my proof, correct? If $dimV_1>1$ its easy. otherwise the dimension is 1. assume for contradcition $dim(T-3I)^2=1$ to get $dimKer(T^2-2T+2I)=dimKer(S)=dimV-1=n-1$. take a basis for that space $B={v_1,...,v_{n-1}}$. and take an eigenvector with eigenvalue $1$, $v_n$. now ${v_1,...,v_n}$ is a basis for V, because if $a_1v_1+...+a_nv_n=0$ implies $a_n = 0$ if we take $S$ of both sides, and 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 ${T-3I,T^2-2T+2I}$. – Omer Apr 6 '19 at 9:24
• @ArturoMagidin 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 $v \in ker(T-3)^2$ and $v \notin V_3$. 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! – Omer Apr 6 '19 at 9:29
• Write it as an answer to your own question; people will have an easier time reading and commenting. – Arturo Magidin Apr 6 '19 at 13:32

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.

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!

• (1) You talk about “$V_3$” but never say what it is. (2) I don’t see why if the dimension is greater than 1, things are easy. You’ll have to do the argument explicitly. (3) $(T-3I)^2$ is a linear transformation; you can’t talk about its dimension. (4) What makes you think there is an eigenvector with eigenvalue 1? On what possible grounds do you claim that? (5) No, you don’t get $(T-3I)^2(v)=0$; why would you get that? – Arturo Magidin Apr 6 '19 at 14:09
• @ArturoMagidin Yes, I guess I wasn't very clear. (1) $V_3$ is the eigenspace of the eigenvalue $3$. (2) If the dimension is greater then 1, then we can have 2 eigenvectors with eigenvalue 3 such that the dimension of their span is 2, and the span is therefore T-Invariant of dimension 2 . (3) yes, i forgot to put the kernel. I'll edit that, the assumption should be dim(ker(T-3I)^2)=1. (4) it should be 3, not sure why I typed 1. I''l edit that too. (5). because v is in the kernel of the transformation (T-3I)^2. thanks for your time – Omer Apr 6 '19 at 14:51
• Yes, this looks right (you are essentially finding a generalized eigenvector of $3$ that is not an eigenvector); not what I was thinking (you can show that a nonzero vector in $\mathrm{range}((T-3I)^2)$ must satisfy $T^2(v) -2T(v)+2v = 0$, and so picking any nonzero vector in that range, $\{v,T(v)\}$ is a 2-dimensional invariant subspace. – Arturo Magidin Apr 6 '19 at 15:16

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 $$span((u+\overline{u}),(u-\overline{u})/i))$$.