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Determinant as a function of trace

  1. Nov 22, 2011 #1
    for dimension 2, the following relation between determinant and trace of a square matrix A is true:

    det A=((Tr A)2-Tr (A2))/2

    for dimension 3 a similar identity can be found in http://en.wikipedia.org/wiki/Determinant

    Does anyone know the generalization to dimension 4 ?

    lukluk
     
  2. jcsd
  3. Nov 22, 2011 #2

    I like Serena

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  4. Nov 22, 2011 #3
    Thanks very much!
    so to see if I understand, the n=4 determinant can be written as

    det A=(p14-6p12p2+3p22+8p1p3-6p4)/24

    where
    pi=Tr (Ai)

    ...right?
     
  5. Nov 22, 2011 #4

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    Right!

    Btw, you can check this yourself if you know a little bit about eigenvalues.

    Did you know that each nxn matrix has n eigenvalues?
    And that the determinant is the product of the eigenvalues?
    And that the trace is the sum of the eigenvalues?
    And the the trace of A^k is the sum of each eigenvalue^k?
     
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