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Math Help - determinant and eigenvalues

  1. #1
    Junior Member
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    determinant and eigenvalues

    For an nxn matrix A

    det(A) = ∏(i=1, .., n) λi ...... [1],

    where the λi are the eigen values (appearing with the correct multiplicities)

    the characteristic equation is det(λI-A)=0, and this is an nth polynomial. Now i can see that statment [1] is true, but not sure how to demontrate this fact. We know that the matrix will be singular if an only if atleast one of the eigenvalues are zero, is there a way i can use this to show that [1] is true? need some advice on how to get started (∏ is from i=1 to n if it was not clear)
    Last edited by CaptainBlack; October 3rd 2006 at 07:53 AM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by action259 View Post
    For an nxn matrix A

    det(A) = ∏(i=1, .., n) λi ...... [1],

    where the λi are the eigen values (appearing with the correct multiplicities)

    the characteristic equation is det(λI-A)=0, and this is an nth polynomial. Now i can see that statment [1] is true, but not sure how to demontrate this fact. We know that the matrix will be singular if an only if atleast one of the eigenvalues are zero, is there a way i can use this to show that [1] is true? need some advice on how to get started (∏ is from i=1 to n if it was not clear)
    If you show that:

    A=PDP^-1,

    where P is the matrix composed of the eigen vectors of A, and D is
    a diaginal matrix with the eigen values of A on the diagonal, then the
    result follows from:

    det(A)=det(P).det(P).det(P^-1),

    and det(P^-1)=1/det(P).

    RonL
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