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,

spans V.

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- May 24th 2010, 07:28 AMChandru1min 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,

spans V. - May 24th 2010, 08:44 AMDefunkt
- May 24th 2010, 09:21 AMChandru1Hi
I think what they mean by v + Av + ... is the subspace generated by

v,Av,A^2v,...,A^{n-1}v - May 25th 2010, 09:02 AMBruno J.
I'd be interested to see a solution to this problem! I tried...

- May 25th 2010, 08:52 PMNonCommAlg
the result is not trivial: i'll take to be the ground field. let be the minimal polynomials of we have because the minimal and characteristic polynomials of are equal.

now for every define see that is an ideal of and thus, since is a PID, is principal, i.e. for some

on the other hand, clearly and so but the number of divisors of is finite, which means the number of distinct elements of the set is finite. let

be those distinct elements and put it is clear that each is a subspace of and therefore for some thus for all

i.e. . that implies by the minimality of and so because we already know that and is monic.

finally, suppose that for some let then , i.e. therefore

which is not possible unless because so we must have for all

**Remark 1**.__second solution__: very briefly, give the structure of an module by defining the multiplication by then see that is a finitely generated torsion module and

thus, since is a PID, we can apply the fundamental theorem for finitely generated modules over PIDs to get some such that now it's easy to show that the set

is linearly independent over

**Remark 2.**the statement in the problem is actually an "if and only if" statement. - May 25th 2010, 09:13 PMBruno J.
Phew!

Awesome proof! (Bow)

(The first one, the one I understand (Smirk)) - May 26th 2010, 08:13 AMChandru1Doubt
Why is ?

- May 26th 2010, 08:30 AMBruno J.