# Thread: min poly = char poly

1. ## min poly = char poly

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,

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

2. Originally Posted by 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,

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

3. ## Hi

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

4. I'd be interested to see a solution to this problem! I tried...

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

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

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

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

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

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

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

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

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

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

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

6. Phew!

Awesome proof!

(The first one, the one I understand )

7. ## Doubt

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

8. Originally Posted by Chandru1
Why is $\Large V = \bigcup\limits_{i=1}^{n} V_{i}$?
Take any $v \in V$; you have $I_v =I_{v_j}$ for some $1 \leq j \leq m$, so $f_{v_j}(A)(v)=0$, so $v \in V_j$.