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Show that for a symmetric or normal matrix

  1. May 5, 2010 #1
    Is there anyway to show that for a symmetric or normal matrix A, that det(A) = [tex]\prod \lambda_i[/tex] without using Jordan blocks? I want to show this result using maybe unitary equivalence and other similar matrices... any ideas? It's obviously easy with JCF...
  2. jcsd
  3. May 5, 2010 #2
    Re: Determinants

    I don't know what is the function of JCF in it.....it simply follows from the well known Caley-Hamilton Theorem (Every square matrix satisfies its own characteristic equation) and the result holds for any square matrix.
  4. May 5, 2010 #3
    Re: Determinants

    Do I have to use Cayley-Hamilton? Could I use the fact that A would be orthogonally equivalent to a diagonal matrix by defintion of symmetric, so for some orthogonal matrix Q and diagonal matrix D:
    then det(A)=det(Q*DQ)=det(D)
    D is diagonal=>det(D)=product of diagonal entries... but how would I show these are the eigenvalues?
    if they're the eigenvalues, then i have my result since similar matrices have the same eigenvalues...
  5. May 6, 2010 #4
    Last edited by a moderator: Apr 25, 2017
  6. May 6, 2010 #5


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    Re: Determinants

    Every symmetric, or normal, matrix, A, can be diagonalized- that is, there exist an invertible matrix P such that [itex]PAP^{-1}= D[/itex] where D is a diagonal matrix having the eigenvalues of A on its diagonal.

    Now [itex]det(PAP^{-1})=[/itex][itex] det(P)det(A)det(P)^{-1}= det(A)= det(D)[/itex] and that last is, of course, the product of the eigenvalues.
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