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Math Help - Prove A is digagonalizablie if...

  1. #1
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    Prove A is digagonalizablie if...

    A is an nxn matrix with two distinct eiganvalues lambda 1 and lambda 2

    the dimension of the eigenspace for lambda 1 is n-1

    prove that A is diagonalizable
    .................................................. ..................................
    so i know the algebraic multiplicity for the first eiganvalue is greater then n-1

    and greater than 1 for the second...

    and for A to diagonalizeable i need the chrasteric polynomial to split...witch it does and also for the eiganspace's dimensions to match they're multiplicty?...if that is correct how to do write it out?
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  2. #2
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    Quote Originally Posted by mtlchris View Post
    A is an nxn matrix with two distinct eiganvalues lambda 1 and lambda 2

    the dimension of the eigenspace for lambda 1 is n-1

    prove that A is diagonalizable
    .................................................. ..................................
    so i know the algebraic multiplicity for the first eiganvalue is greater then n-1

    and greater than 1 for the second...

    and for A to diagonalizeable i need the chrasteric polynomial to split...witch it does and also for the eiganspace's dimensions to match they're multiplicty?...if that is correct how to do write it out?
    "for A to diagonalizeable i need the chrasteric polynomial to split...witch it does" ..... how do you say that?

    As you said,
    "so i know the algebraic multiplicity for the first eiganvalue is greater then n-1......and greater than 1 for the second..."

    The algebraic multiplicity of the first eigenvalue cannot be n because the second eigenvalues algebraic multiplicity is at least 1. Thus the algebraic multiplicity of the first eigenvalue is exactly n-1. This means the algebraic multiplicity and the geom. multiplicity of the second eigenvalue is exactly 1 and thus the matrix is diagonalizable.
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  3. #3
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    I say that the characteristic polynomial splits because A has two distinct eigenvalues so the characteristic polynomial has to be a product of linear factors...right??
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