1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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;
  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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook