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Trace and derivatives

  1. Mar 30, 2010 #1
    Hey guys, any hints on how to show that [tex]\frac{d}{dt}[/tex]|t=0 [tex]det(I + tA) = tr(A) [/tex]? I did it for 2x2 but I can't figure out a generalization. Thanks
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  3. Mar 30, 2010 #2


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    Did you try to use the definition of the determinant to conclude something?
  4. Mar 30, 2010 #3
    Which definition? Hmm, the co-factor expansion?
  5. Mar 30, 2010 #4


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    Some people have a thing about Wikipedia articles, but I don't believe it matters right now.


    So, after examining this definition (which may seem a bit complicated at the first glance, and if there is an easier way to prove it, it would be nice) you can conclude that the only permutation which generates a polynomial with a non-zero coefficient for t is the identity permutation. After taking the derivative with respect to t of your determinant, and setting t = 0, everything vanishes except the coefficient of t, which equals Tr(A). This can be seen if you try to multiply (t a11 + 1)(t a22 + 1), or (t a11 + 1)(t a22 + 1)(t a33 + 1), etc. i.e. you will always have a term of the form t(a11 + a22 + a33 + ...) generated. All the other polynomials appearing in your sum of permutations are irrelevant, since they all vanish at t = 0, and the constant term is eliminated by taking the derivative itself.
  6. Mar 30, 2010 #5


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    Oh yes, and note that the identity permutation is an even one.
  7. Mar 30, 2010 #6
    Note that [itex]\det \left(I+tA\right)=\det \left(I-\left(-t\right)A\right)[/itex] is the characteristic polynomial of [itex]A[/itex] evaluated at -t.

    Now, for any polynomial, the t term is the symmetric of the sum of its roots; you may see this by looking at its factorization:


    So, when you differentiate [itex]p\left(-t)[/itex] and set [itex]t=0[/itex], what will be the only term left? And how is the trace of [itex]A[/itex] related to its eigenvalues?
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