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Dr Euler and characteristic equations

  1. Feb 7, 2007 #1
    In "Dr. Euler's Fabulous Formula" by Paul Nahin, early in chapter 1 is discussed characteristic polynomials of a square matrix and the Cayley-Hamilton theorem, that any square matrix A satisfies its own characteristic equation. On page 21 it states p(lambda) = lambda^2 + a1*lambda + a2 = 0 and then goes on to say "suppose we divide lambda^n by lambda^2 + a1*lambda + a2. The most general result is a polynomial of degree n - 2 and a remainder of at most degree one." Apparently this is an important result, but isn't this also an invalid divide by zero? Am I misunderstanding something here? I checked out characteristic equations in Schaum's Outline on Matrices but it doesn't touch on this point. Can somebody explain this? Thanks.
     
  2. jcsd
  3. Feb 7, 2007 #2

    matt grime

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    That is not divide by zero. You're dividing a (definitely non-zero) polynomial into another. Remember to divide f by g in this sense means to find polys q and r with

    f=qg+r

    and deg(r)<deg(f).
     
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