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Math Help - minimal polynomials

  1. #1
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    minimal polynomials

    Let V be a finite dimensional vector space over the field F, with dim(V)=n. We know that End(V), the space of linear transformations from V to itself. dim(End(V)) is n^2.

    The minimal polynomial of a linear transformation T is the unique monic polynomial which generates the ideal Ann(V) in F[x]. This polynomial divides the characteristic polynomial of T. Since the degree of the characteristic polynomial is n, the minimal polynomial is at most degree n.

    I am asked to prove that the minimal polynomial of T has at most degree n^2, using that End(V) has dimension n^2. So I am asked to prove a weaker bound on the degree than I already know is true.
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  2. #2
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    Quote Originally Posted by robeuler View Post
    Let V be a finite dimensional vector space over the field F, with dim(V)=n. We know that End(V), the space of linear transformations from V to itself. dim(End(V)) is n^2.

    The minimal polynomial of a linear transformation T is the unique monic polynomial which generates the ideal Ann(V) in F[x]. This polynomial divides the characteristic polynomial of T. Since the degree of the characteristic polynomial is n, the minimal polynomial is at most degree n.

    I am asked to prove that the minimal polynomial of T has at most degree n^2, using that End(V) has dimension n^2. So I am asked to prove a weaker bound on the degree than I already know is true.
    I think the idea here is to let A be the matrix corresponding to T, then \{ I,A,A^2,...,A^{n^2}\} is linearly dependent because it has n^2+1 elements. Therefore, there exists a_j \in F, 0\leq j\leq n^2+1 so that \sum_{j=0}^{n^2+1}a_j A^j = \bold{0}. Therefore, f(X) = \sum_{j=0}^{n^2+1} a_j\cdot I ~X^j is a polynomial which has A as a zero (I is the identity matrix).
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