# min poly = char poly

#### Chandru1

Let A:V->V be Linear Transformation. dim V =n. Prove that if Characteristic polynomial of A = min. Polynomial A then, there is a v \in V such that,

$$\displaystyle v+ Av+ \cdots A^{n-1}v$$ spans V.

#### Defunkt

MHF Hall of Honor
Let A:V->V be Linear Transformation. dim V =n. Prove that if Characteristic polynomial of A = min. Polynomial A then, there is a v \in V such that,

$$\displaystyle v+ Av+ \cdots A^{n-1}v$$ spans V.
Surely, you mean that $$\displaystyle \{v,Av,...,A^{n-1}v\}$$ span $$\displaystyle V$$, yes? One vector can't span a space of dimension >1...

#### Chandru1

Hi

I think what they mean by v + Av + ... is the subspace generated by
v,Av,A^2v,...,A^{n-1}v

#### Bruno J.

MHF Hall of Honor
I'd be interested to see a solution to this problem! I tried...

#### NonCommAlg

MHF Hall of Honor
I'd be interested to see a solution to this problem! I tried...
the result is not trivial: i'll take $$\displaystyle F$$ to be the ground field. let $$\displaystyle p(x) \in F[x]$$ be the minimal polynomials of $$\displaystyle A.$$ we have $$\displaystyle \deg p(x)=n$$ because the minimal and characteristic polynomials of $$\displaystyle A$$ are equal.

now for every $$\displaystyle v \in V$$ define $$\displaystyle I_v= \{f(x) \in F[x]: \ f(A)(v) = 0 \}.$$ see that $$\displaystyle I_v$$ is an ideal of $$\displaystyle F[x]$$ and thus, since $$\displaystyle F[x]$$ is a PID, $$\displaystyle I_v$$ is principal, i.e. $$\displaystyle I_v= \langle f_v(x) \rangle,$$ for some $$\displaystyle f_v(x) \in F[x].$$

on the other hand, clearly $$\displaystyle p(x) \in I_v$$ and so $$\displaystyle f_v(x) \mid p(x).$$ but the number of divisors of $$\displaystyle p(x)$$ is finite, which means the number of distinct elements of the set $$\displaystyle \{I_v: \ v \in V \}$$ is finite. let $$\displaystyle I_{v_1}, \cdots , I_{v_m}$$

be those distinct elements and put $$\displaystyle V_i = \{v \in V: \ f_{v_i}(A)(v)=0 \}.$$ it is clear that each $$\displaystyle V_i$$ is a subspace of $$\displaystyle V$$ and $$\displaystyle V=\bigcup_{i=1}^m V_i.$$ therefore $$\displaystyle V=V_k,$$ for some $$\displaystyle 1 \leq k \leq m.$$ thus $$\displaystyle f_{v_k}(A)(v)=0,$$ for all

$$\displaystyle v \in V,$$ i.e. $$\displaystyle f_{v_k}(A)=0$$. that implies $$\displaystyle p(x) \mid f_{v_k}(x),$$ by the minimality of $$\displaystyle p(x),$$ and so $$\displaystyle f_{v_k}(x)=p(x)$$ because we already know that $$\displaystyle f_{v_k}(x) \mid p(x)$$ and $$\displaystyle p(x)$$ is monic.

finally, suppose that $$\displaystyle c_0v_k + c_1A(v_k) + \cdots + c_{n-1}A^{n-1}(v_k)=0,$$ for some $$\displaystyle c_j \in F.$$ let $$\displaystyle g(x)=c_0 + c_1x + \cdots + c_{n-1}x^{n-1} \in F[x].$$ then $$\displaystyle g(A)(v_k)=0$$, i.e. $$\displaystyle g(x) \in I_{v_k} = \langle f_{v_k}(x) \rangle = \langle p(x) \rangle.$$ therefore

$$\displaystyle p(x) \mid g(x),$$ which is not possible unless $$\displaystyle g = 0,$$ because $$\displaystyle \deg p(x)=n > \deg g(x).$$ so we must have $$\displaystyle c_j = 0,$$ for all $$\displaystyle j. \ \Box$$

Remark 1. second solution: very briefly, give $$\displaystyle V$$ the structure of an $$\displaystyle F[x]$$ module by defining the multiplication by $$\displaystyle f(x)v = f(A)(v).$$ then see that $$\displaystyle V$$ is a finitely generated torsion $$\displaystyle F[x]$$ module and

thus, since $$\displaystyle F[x]$$ is a PID, we can apply the fundamental theorem for finitely generated modules over PIDs to get some $$\displaystyle v \in V$$ such that $$\displaystyle \text{ann}_{F[x]}(v)=\text{ann}_{F[x]}(V).$$ now it's easy to show that the set

$$\displaystyle \{v,A(v), \cdots , A^{n-1}(v) \}$$ is linearly independent over $$\displaystyle F.$$

Remark 2. the statement in the problem is actually an "if and only if" statement.

Last edited:
• Defunkt and Bruno J.

#### Bruno J.

MHF Hall of Honor
Phew!

Awesome proof! (Bow)

(The first one, the one I understand (Smirk))

#### Chandru1

Doubt

Why is $$\displaystyle \Large V = \bigcup\limits_{i=1}^{n} V_{i}$$?

#### Bruno J.

MHF Hall of Honor
Why is $$\displaystyle \Large V = \bigcup\limits_{i=1}^{n} V_{i}$$?
Take any $$\displaystyle v \in V$$; you have $$\displaystyle I_v =I_{v_j}$$ for some $$\displaystyle 1 \leq j \leq m$$, so $$\displaystyle f_{v_j}(A)(v)=0$$, so $$\displaystyle v \in V_j$$.