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Math Help - min poly = char poly

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by Chandru1 View Post
    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...
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  3. #3
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    Hi

    I think what they mean by v + Av + ... is the subspace generated by
    v,Av,A^2v,...,A^{n-1}v
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    MHF Contributor Bruno J.'s Avatar
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    I'd be interested to see a solution to this problem! I tried...
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    Quote Originally Posted by Bruno J. View Post
    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.
    Last edited by NonCommAlg; May 25th 2010 at 10:32 PM.
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  6. #6
    MHF Contributor Bruno J.'s Avatar
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    Phew!

    Awesome proof!

    (The first one, the one I understand )
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  7. #7
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    Doubt

    Why is \Large V = \bigcup\limits_{i=1}^{n} V_{i}?
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  8. #8
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by Chandru1 View Post
    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.
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