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Eigenvalue show that

  1. Dec 6, 2006 #1
    eigenvalue "show that"

    1. The problem statement, all variables and given/known data
    Let A be a matrix whose columns all add up to a fixed constant [tex]\delta[/tex]. Show that [tex]\delta[/tex] is an eigenvalue of A

    2. Relevant equations

    3. The attempt at a solution
    My solution manual's hint is: If the columns of A each add up to a fixed constant [tex]\delta[/tex], then the row vectors of [tex]A - \delta I[/tex] all add up to (0,0....0).

    I don't even understand the hint.
  2. jcsd
  3. Dec 6, 2006 #2
    First of all do you understand why "If the columns of A each add up to a fixed constant [tex]\delta[/tex], then the row vectors of [tex]A - \delta I[/tex] all add up to (0,0....0)."?

    If yes, then

    What is the equation that [tex]delta[/tex] has to fit in order to be an eigenvalue of [tex]A[/tex]?

    What is the relation between the determinant of matrix [tex]A[/tex] and the determinant of the matrix obtained by adding to one of the rows of matrix [tex]A[/tex] all the others?

    What is the determinant of a matrix that has a row of 0's?
    Last edited: Dec 6, 2006
  4. Dec 6, 2006 #3
    1. Ax = lambda*x ?

    2. det(A)

    3. 0.

  5. Dec 6, 2006 #4
    Yes but more helpful det([tex]A - \delta I[/tex]) = 0
  6. Dec 6, 2006 #5
    so obviously I see the answer IF i can show that somehow I can get A to include a row of all zeroes.
  7. Dec 6, 2006 #6
    Ok. Because each column of A adds up to a fixed constant [tex]\delta[/tex], it means that the rows (and the columns) of A add up to a constant of n*[tex]\delta[/tex], which means that the rows of [tex]A - \delta I[/tex] add up to 0.
    So the matrix formed by, say, adding to the first row of [tex]A - \delta I[/tex] all the other rows will have the first row all 0's, and the same determinant as [tex]A - \delta I[/tex], which means that det([tex]A - \delta I[/tex]) = 0 and [tex]\delta[/tex] is an eigenvalue of A
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