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Math Help - Eigenvalue

  1. #1
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    Eigenvalue

    Let \lambda be an eigenvalue of a square matrix A.

    How to show that:

    1. \lambda^n is an eigenvalue of A^n where n is a positive integer.

    2. If A is invertible, show that \frac {1}{\lambda} is an eigenvalue of A^{-1}.

    To start off: I was told to let x be an eigenvector such that Ax= \lambda x
    ....
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  2. #2
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    Hello,
    Quote Originally Posted by noob mathematician View Post
    Let \lambda be an eigenvalue of a square matrix A.

    How to show that:

    1. \lambda^n is an eigenvalue of A^n where n is a positive integer.
    If \lambda is an eigenvalue of A, then there exists X such that :
    AX=\lambda X

    Now multiply to the left by A :
    A^2X=A \lambda X
    Since \lambda is a scalar, it commutes with A.

    Thus A^2X=\lambda AX=\lambda (\lambda X)=\lambda^2 X
    Hence \lambda^2 is an eigenvalue of A^2

    You can use this process to prove your question by induction.

    2. If A is invertible, show that \frac {1}{\lambda} is an eigenvalue of A^{-1}.
    AX=\lambda X
    Multiply to the left by A^{-1} :

    I_nX=X=\lambda A^{-1}X

    And same reasoning as above.
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  3. #3
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    Quote Originally Posted by noob mathematician View Post
    Let \lambda be an eigenvalue of a square matrix A.

    How to show that:

    1. \lambda^n is an eigenvalue of A^n where n is a positive integer.

    2. If A is invertible, show that \frac {1}{\lambda} is an eigenvalue of A^{-1}.

    To start off: I was told to let x be an eigenvector such that Ax= \lambda x
    ....
    Part 1 is very easy, just follow the hint. Ax = \lambda x so A^2 x  \  = \   A \lambda x \   = \  \lambda A x  = \lambda^2 x. it should be obvious that \lambda^n<br />
is an eigenvalue of A^n you may want to write down an inductive argument.

    part 2 also doesn't present much difficulty. Ax = \lambda x \rightarrow  A^{-1} A x =  A^{-1} \lambda x so  A^{-1} \lambda x = x.

    Bobak

    Edit: i was beaten where had the delete post button gone ?
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  4. #4
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    ...it should be obvious that
    \lambda^n is an eigenvalue of A^n...
    Do that imply that A^n will always have the same eigenvalues as A^1
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  5. #5
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    Quote Originally Posted by aidan View Post
    Do that imply that A^n will always have the same eigenvalues as A^1
    no, just the same eigenvectors.

    Bobak
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  6. #6
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    Quote Originally Posted by bobak View Post
    no, just the same eigenvectors.

    Bobak
    I did not phrase that question correctly.
    What I meant:

    if  A^1 has eigenvalues
     \lambda_1 \\ \lambda_2 \lambda_3

    then will  A^n have eigenvalues
     \lambda_1^n \lambda_2^n \lambda_3^n

    Based on the answer given above to the original question, it just seems that if it works for one eigenvalue, it will work for all.

    When a matrix is raised to a power, will the eigenvalues also be raised to the same power?
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  7. #7
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    Quote Originally Posted by aidan View Post
    I did not phrase that question correctly.
    What I meant:

    if  A^1 has eigenvalues
     \lambda_1 \\ \lambda_2 \lambda_3

    then will  A^n have eigenvalues
     \lambda_1^n \lambda_2^n \lambda_3^n

    Based on the answer given above to the original question, it just seems that if it works for one eigenvalue, it will work for all.

    When a matrix is raised to a power, will the eigenvalues also be raised to the same power?
    Yes (was anything special assumed about the eigenvalue in your original post ....?)
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