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Math Help - eigenvalue/eigenvector

  1. #1
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    eigenvalue/eigenvector

    Show that if a matrix A has an eigenvector x corresponding to the eigenvalue h, then x is also an eigenvector of the matix A^2 and state the corresponding eigenvalue.
    Write an analogous result for A^n.

    I'm afraid this kind of has me stumped. I have managed to do a couple of questions on eigenvalues and eigenvectors, but I can't apply that knowledge to this problem.

    Anyone help?!?
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  2. #2
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    If k is an eigenvalue of A then it means A\bold{x} = k\bold{x} has a non-trivial solution \bold{x}.
    But that means A^2\bold{x}=A(A\bold{x})=A(k\bold{x}) = k(A\bold{x}) = k(k\bold{x})=k^2\bold{x}.
    Thus, \bold{x} is an eigenvector of A^2 corresponding with eigenvalue k^2.
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  3. #3
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    Thank you for the help ThePerfectHacker !

    I am wondering if someone can help me with another similar problem. Rather than start a new thread I'll ask here. Please have mercy, I am not sure how to format the math correctly. Here goes...

    I have a matrix:
    1 -1
    1 0

    I have got the eigenvalues to be: (1 +/- sqrt3) / 2
    or e^(i.pi/3) & e^-(i.pi/3)

    I must now show what the corresponding eigenvectors are, and I am not sure of the answer.
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  4. #4
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by zbrn View Post
    Thank you for the help ThePerfectHacker !

    I am wondering if someone can help me with another similar problem. Rather than start a new thread I'll ask here. Please have mercy, I am not sure how to format the math correctly. Here goes...

    I have a matrix:
    1 -1
    1 0

    I have got the eigenvalues to be: (1 +/- sqrt3) / 2
    or e^(i.pi/3) & e^-(i.pi/3)

    I must now show what the corresponding eigenvectors are, and I am not sure of the answer.
    let k be an eigenvalue of the matrix A..
    then the corresponding eigenvector are the matrices of the form

    (kI - A) X = 0 and solve for the x's..
    note, they may not be unique. in fact, they form a set called the Eigenspace with respect to k.. (note again that as the name implies, the set is a space..)
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