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Math Help - Generalized eigenspace

  1. #1
    Newbie erich22's Avatar
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    Generalized eigenspace

    Can someone help?
    1. Is a generalized eigenspace an eigenspace?
    2. How can we be sure that a generalized eigenspace has a full rank?
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  2. #2
    MHF Contributor

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    Re: Generalized eigenspace

    Quote Originally Posted by erich22 View Post
    Can someone help?
    1. Is a generalized eigenspace an eigenspace?
    2. How can we be sure that a generalized eigenspace has a full rank?
    regarding your second question, i'm not sure what you mean by a "full rank eigenspace". so i'll wait until you define that term.

    your first question is easy to answer though:

    suppose that V is a finite dimensional vector space, T an operator on V, \lambda an eigenvalue of T and W the generalized eigenspace corresponding to \lambda.

    suppose that W is an eigenspace. then there exists a scalar \mu such that W = \{x \in V: \ Tx = \mu x \}. now if (T-\lambda I)^k x = 0, for some k \geq 1 and

    x \neq 0, then \sum_{i+j=k} (-1)^i \binom{k}{i}\lambda^i T^j(x)=0 and, since T^j(x)=\mu^j x, we'll get \sum_{i+j=k}(-1)^i \binom{k}{i}\lambda^i \mu^jx = 0. thus (\mu - \lambda)^kx = 0 which implies \mu = \lambda.

    so the only way that a generalized eigenspace becomes an eigenspace is just the trivial one.

    for example, consider the linear transformation T: \mathbb{C}^2 \longrightarrow \mathbb{C}^2 defined by T(x,y)=(x,0). then the generalized eigenspace corresponding to \lambda = 0 is the eigenspace corresponding to \lambda=0.
    Last edited by NonCommAlg; September 3rd 2011 at 10:21 PM.
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  3. #3
    Newbie erich22's Avatar
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    Re: Generalized eigenspace

    Thank u so much for the 1st answer. It really2 helps.
    For the second question, I mean how can we be sure that a generalized eigenspace W has the same dimension as the algebra multiplicity of the eigenvalue,lambda?
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  4. #4
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    Re: Generalized eigenspace

    Every matrix satisfies its own characteristic equation. That is, if the characteristic equation for matrix A is (\lambda- a)^n= 0 then (A- aI)^nv= 0 for every v. It might happen that there are n independent eigenvectors in which case the "eigenspace" has dimension n. If not, if the eigenspace has dimension m< n, for v NOT in that eigenspace, we must have (A- aI)^(n-m)v= 0 so that v is a "generalized" eigen value.
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