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Homework Help: Possible dimensions of eigenspaces, known characteristic polynomial

  1. Jul 12, 2008 #1
    1. The problem statement, all variables and given/known data
    If A is a 6x6 matrix with characteristic polynomial:
    what are the possible dimensions of the eigenspaces?

    3. The attempt at a solution

    The solution given is that, for each each eigenspace, the smallest possible dimension is 1 and the largest is the multiplicity of the eigenvalue (the number of times the root of the characteristic polynomial is repeated). So, for the eigenspace corresponding to the eigenvalue 2, the dimension is 1, 2, or 3.

    I do not understand where this answer comes from. Please come someone help me understand this.
  2. jcsd
  3. Jul 12, 2008 #2
    Re: Eigenspaces

    Since you are given a square matrix i.e. 6x6 matrix and the characteristic polynomial
    What are the eigenvalues? What are the multiplicities of each eigenvalue?

    What can you say about the dimension for the eigenspace of the eigenvalues? it is "blank" than the multiplicity of the eigenvalue [tex]\lambda_{k}[/tex] for [tex]1 \leq k \leq p[/tex] where p is the number of distinct eigenvalues, which is basically what the book is telling you.
    Last edited: Jul 12, 2008
  4. Jul 12, 2008 #3
    Re: Eigenspaces

    Yes, I understand that. That was me axplaining the answer. I'm asking how they got to that answer. Sorry about the ambiguity.

    If an eigenvalue has multiplicity n in the characteristic polynomial, why should its eigenspace have dimension between 1 and n? That's what I don't understand, why is that the case?
  5. Jul 12, 2008 #4
    Re: Eigenspaces

    The eigenspace is the null space(kernel) of the matrix [tex]A- \lambda I[/tex], which is subspace of [tex]R^{n}[/tex].
    By definition of a dimension of a non-zero subspace, it is the number of independent vectors in the basis for the eigenspace.
    The multiplicity of the eigenvalue would then be the maximum number, n, of linearly independent vectors in the subspace. Since the eigenspace is non-zero then the dimension must be greater or equal to 1 and the maximum number of independent vectors in the basis is n. If n=3 when [tex]\lambda = 2[/tex], then the dimension can be one-dimensional, two-dimensional, or three-dimensional.
  6. Jul 12, 2008 #5


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    Science Advisor

    Re: Eigenspaces

    From knowing that the characteristic equation is x2(x-1)(x-2)3, you immediately know that the eigenvalues are 0, 1, and 2. Since the x-1 term is linear, you know that the dimension of the subspace of eigenvectors with eigenvalue 1 must be of dimension 1. Since x2 term is of degree two, you know that the subspace of eigenvectors with eigenvalue 1 has dimension either 1 or 2. Since (x-2)3 is of degree 3, you know that the subspace of eigenvectors with eigenvalue 2 has dimension 1, 2, or 3.
  7. Jul 14, 2008 #6
    Re: Eigenspaces

    I'm feeling a bit thick. Please help me connect the dots. (x-2)^3 is cubic, and P_3 has dimension four? Why can't a cubic term have dimension 4? But it can have dimension less than 3?

    I just don't make th connexion from knowing the degree of the term in the characteristic polynomial, and deducing the dimension of the corresponding eigenspace.
  8. Jul 15, 2008 #7

    matt grime

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    Homework Helper

    Re: Eigenspaces

    Just think about the Jordan Normal form (or take it as a definition).
  9. Jul 17, 2008 #8
    Re: Eigenspaces

    Sorry, I haven't done that (Jordan Normal form) yet.
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