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Eigenfunctions & Eigenvalues

  1. Mar 27, 2008 #1
    a) Consider a linear operator L with 2 different eigenvalues a1 and a2, with their corresponding eigenfunction f1 and f2. Is f1 + f2 also an eigenfunction of L? If so, what eigenvalue of L does it correspond to? If not, why not?

    b) Answer the same question as in part (a) but for the difference of the 2 functions;
    f1-f2.
     
  2. jcsd
  3. Mar 27, 2008 #2
    Let's see:

    L(f1 + f2) = L(f1) + L(f2) (because L is linear)

    = a1 f1 + a2 f2

    If f1 + f2 is an eigenfunction then we must have:

    L(f1 + f2) = b (f1 + f2)

    for some eigenvalue b. This means that:

    a1 f1 + a2 f2 = b f1 + b f2 ------>

    (a1 - b) f1 + (a2 - b) f2 = 0

    which means that f1 and f2 are proportional to each other. However, that is impossible because then the iegenvalues a1 and a2 have to bethe same. So, we arrive at a contradiction and f1 + f2 cannot be an eigenfunction of L.

    Part b) minus f2 is also an eigenfunction with eigenvalue a2. So, the result of a) also applies in this case.
     
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