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Prove that every non-zero vector in V is a maximal vector

  1. Nov 19, 2012 #1
    1. The problem statement, all variables and given/known data
    Let V be a finite dimensional vector space and T is an operator on V. Assume [itex]μ_{T}(x)[/itex] is an irreducible polynomial. Prove that every non-zero vector in V is a maximal vector.

    2. Relevant equations
    [itex]μ_{T}(x)[/itex] is the minimal polynomial on V with respect to T.

    3. The attempt at a solution
    I am not sure how to go about solving this problem. I know that if T is cyclic then [itex]\exists \vec{v} \in V[/itex] such that [itex]<T,\vec{v}>=V[/itex], then for any [itex]\vec{w}\in V[/itex],
    [itex]\exists f(x) \in F[x][/itex] such that [itex] \vec{w} = f(T)(\vec{v})[/itex]
    and it follows that [itex]μ_{T,\vec{w}}(x) = \frac{μ_{T}(x)}{gcd(μ_{T}(x),f(x))}[/itex]
    However, since [itex]μ_{T}(x)[/itex] is an irreducible polynomial, then [itex]gcd(μ_{T}(x),f(x)) = μ_{T}(x)[/itex] or [itex]gcd(μ_{T}(x),f(x)) = 1[/itex]
    If [itex]gcd(μ_{T}(x),f(x)) = μ_{T}(x)[/itex], then [itex]μ_{T,\vec{w}}(x) = 1[/itex], which implies that [itex]\vec{w} = I(\vec{w}) = \vec{0}[/itex]
    If [itex]gcd(μ_{T}(x),f(x)) = 1[/itex], then [itex]μ_{T,\vec{w}}(x) = μ_{T}(x)[/itex], which means that [itex]\vec{w}[/itex] is a maximal vector. Thus every non-zero vector in V is a maximal vector.

    However, I'm not sure how to prove that T is cyclic.

    I also attempted to find the minimal polynomials with respect to the basis vectors for V.
    Since [itex]μ_{T}(x) = lcm(μ_{T,\vec{v_1}}(x),μ_{T\vec{v_2}}(x),...,μ_{T,\vec{v_n}}(x))[/itex]
    [itex]\Rightarrow μ_{T}(x)=μ_{T,\vec{v_1}}(x)=μ_{T,\vec{v_2}}(x)=...=μ_{T,\vec{v_n}}(x)[/itex] since the minimal polynomial is irreducible. But then I didn't know where to go from there either.
     
  2. jcsd
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