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Determinants and taylor expansion

  1. Sep 17, 2013 #1
    I'm doing a proof, and near the last step I want to write the expression,

    [tex] \frac{d}{dt} \det{A(t)} = \lim_{\epsilon \to 0} \frac{\det{(A+\epsilon \frac{dA}{dt})} - \det{A}}{\epsilon} [/tex]

    which produces the right answer, so I believe that it may be correct. This looks very much like a Taylor expansion, but I'm having trouble justifying it exactly - does anyone know the proper way to Taylor expand a determinant, and if the above expression is actually true?
     
  2. jcsd
  3. Sep 17, 2013 #2
    What you are really doing is: ## \det A(t + \epsilon) = \det (A(t) + B(t, \epsilon)) ##, where ##B(t, \epsilon) = A(t + \epsilon) - A(t) ##, which is a matrix. Show that as ## \epsilon \to 0 ##, ##B(t, \epsilon) \to A'(t)\epsilon ##.
     
  4. Sep 17, 2013 #3
    Ah yes I see now. The old "add zero" trick. ;)

    Thank you!
     
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