# Eigenfunctions & Eigenvalues

1. Mar 27, 2008

### eit32

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. Mar 27, 2008

### Count Iblis

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.