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Minimal Polynomial A nxn Matrix

  1. Oct 26, 2006 #1
    Let A be an n x n matrix; denote its distinct eigenvalues by a_1,...,a_k and denote the index of a_i by d_i. How do I prove that the minimal polynomial is then:

    m_A(s) = (s-a_1)^d_1*...*(s-a_k)^d_k

    ?

    The characterstic polynomial is defined as:

    p_A(s) = (s-a_1)*...*(s-a_n);
     
    Last edited: Oct 26, 2006
  2. jcsd
  3. Oct 26, 2006 #2

    Hurkyl

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    What do you mean by index? If you mean "multiplicity", then you can't prove it. For example, consider the minimum polynomials of the zero and identity matrices. (1 and x-1, respectively)
     
  4. Oct 26, 2006 #3
    The book defines the index as:

    "N_m = N_m(a) the nullspace of (A-a*I)^m. The subspaces N_m consist of generalized eigenvectors; they are indexed increasingly, that is N_1 is included in N_2 is included in...

    Since these are subspaces of a finite-dimensional space, they must be equal from a certain index on. We denote by d = d(a) the smallest such index, that is, N_d = N_(d+1) = ... but N_(d-1) /= N_d_j

    d(a) is called the index of the eigenvalue a."

    This is from Peter D. Lax's book on linear algebra.
     
    Last edited: Oct 26, 2006
  5. Oct 26, 2006 #4

    Hurkyl

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    If f is the minimal polynomial of A, then f(A) is the zero matrix. So f(A)v = 0 for any vector. Does that help?
     
  6. Oct 26, 2006 #5
    That is also true for the characteristic polynomial. I know that the characteristic polynomial can be divided by a minimal polynomial and I want to show that the minimal polynomial is equal to (s-a_1)^(d_1)*...*(s-a_k)^(d_k)
     
  7. Oct 26, 2006 #6

    Hurkyl

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    Right. But you want the smallest polynomial that satisfies what I wrote. So you first want to prove:

    (s-a_1)^(d_1)*...*(s-a_k)^(d_k)

    satisfies what I wrote, and then you want to prove that any proper factor of it does not. I think good choices of f(A)v will do that.
     
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