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Math Help - tests coming up-eigenvectors-prove

  1. #1
    n22
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    tests coming up-eigenvectors-prove

    Hi ,
    I would really appreciate if someone could help me out here.thanks.

    Prove that if an nxn matrix A has n distinct eigenvalues ,then the set of n eigenvectors formed by
    choosing one non zero vector from each eigenspace of A is a basis of R^n
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  2. #2
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    Re: tests coming up-eigenvectors-prove

    Hey n22.

    Are you familiar with the diagonalization theorem?

    Diagonalizable matrix - Wikipedia, the free encyclopedia
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    n22
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    Re: tests coming up-eigenvectors-prove

    Quote Originally Posted by chiro View Post
    Hey n22.

    Are you familiar with the diagonalization theorem?

    Diagonalizable matrix - Wikipedia, the free encyclopedia
    Hello,
    I am familiar with the theorem. but knowing how to prove something is difficult-especially when I dont understa"choosing one non zero vector from each eigenspace" is .So if I picked any two vectors then its a basis of R^n.?please expand.thanks.
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    Re: tests coming up-eigenvectors-prove

    Quote Originally Posted by n22 View Post
    Hello,
    I am familiar with the theorem. but knowing how to prove something is difficult-especially when I dont understa"choosing one non zero vector from each eigenspace" is .So if I picked any two vectors then its a basis of R^n.?please expand.thanks.
    Not "any two vectors", one vector from each eigenspace. Do you know what an eigenspace is? You should have already proved that if v satisfies Av= \lamba v then so does any multiple of it: A(av)= a(Av)= a(\lambda v)= \lambda (av). From that it follows that the set of all eigenvectors, corresponding to a given eigenvalue (together with the 0 vector) is a subspace.

    It is also easy to see that if two eigenvectors correspond to different eigenvalues, they are independent (one is not a multiple of the other). So if A has n different eigenvalues, I has n different eigenspaces.
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