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Math Help - determinant, eigenvalue

  1. #1
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    determinant, eigenvalue

    How do you find the left eigenvalue for a matrix where P is an N xN matric where all its entries are 1/N?

    i know that i need to use characteristic polynomial but how do i find the determinant for it? its too tedious by this method.
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  2. #2
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    Re: determinant, eigenvalue

    suppose that A is the matrix in question. we have A = (1/N)B, where B is a matrix with all 1's.

    so det(A-xI) = det((1/N)B - xI) = det((1/N)(B - xNI) = det((1/N)(B - NxI)) = ((1/N)^N)(det(B - NxI)).

    if we let y = Nx, we have (1/N)^N(det(B - yI)).

    so let's look at that determinant, det(B-yI) =

    \begin{vmatrix}1-y&1&\dots&1\\1&1-y&\dots&1\\ \vdots&\vdots&\ddots&\vdots\\1&1&\dots&1-y\end{vmatrix}

    if we subtract the first row from every other row (which does not change the determinant), we get:

    \begin{vmatrix}1-y&1&\dots&1\\y&-y&\dots&0\\ \vdots&\vdots&\ddots&\vdots\\y&0&\dots&-y\end{vmatrix}

    if we add every column to the first (which also does not change the determinant), we get:

    \begin{vmatrix}N-y&1&\dots&1\\0&-y&\dots&0\\ \vdots&\vdots&\ddots&\vdots\\0&0&\dots&-y\end{vmatrix}

    can you find the determinant of this last matrix? this should give you the only possible eigenvalues rather easily.
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Re: determinant, eigenvalue

    Quote Originally Posted by alexandrabel90 View Post
    How do you find the left eigenvalue for a matrix where P is an N xN matric where all its entries are 1/N?

    i know that i need to use characteristic polynomial but how do i find the determinant for it? its too tedious by this method.
    You obtain a triangular determinant with the following steps:

    (i)\;\; R_2-R_1,\ldots,R_n-R_1\quad (ii)\;\;C_1+C_2+\ldots+C_n

    Edited: Sorry, I didn't see Deveno's post.
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  4. #4
    Super Member ILikeSerena's Avatar
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    Re: determinant, eigenvalue

    Hi alexandrabel90!

    Your matrix P has N identical columns.
    This means that the column space is 1-dimensional.
    In turn this means that the null space is (N-1)-dimensional.
    Therefore 0 is an eigenvalue of P with multiplicity (N-1).

    So you have only 1 other eigenvalue that is not zero.
    The trace of P is the sum of the eigenvalues.
    Can you guess what the value of this last eigenvalue must be?


    Edit: Ah, it seems we have been stacking replies.
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  5. #5
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    Re: determinant, eigenvalue

    therefore def(B-yI) = (N-y) (-y)^(N-1)
    and thus det (A-xI) = (N-Nx) (-Nx)^ (N-1) (1/N)^(N)

    if i let this equal to 0, how do i find the value that x can take given that N is unknown?
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  6. #6
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    Re: determinant, eigenvalue

    y can only be 0 or N. what does that mean x can equal, given that y = Nx?
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