Eigenvalues of an operator I think this question isn't that hard, but I am a bit confused:
Define $$(Af)(x):=\int_{0}^{1}\cos(2\pi(x-y))f(y)dy.$$ Then $A$ is an operator on functions.  Find the eigenvalues and the eigenfunctions.
I can think of a lot of functions that give $0$, things like $f(x)=\cos(n2\pi x)$.  Also one eigenfunction that gives eigenvalue $\frac {1}{2}$ (I think).  My problem is I have no idea how to show that I have found all of them, and I don't know if I have found all of them.
Thanks for showing me!!
Edit:  I havent really shown my work because there is a lot of it and it is all over the place.  J.M. suggest that it is useful to write $$Af(x)=\cos(2\pi x)\int_{0}^{1}\cos\left(2\pi y\right)f(y)dy+\sin\left(2\pi x\right)\int_{0}^{1}\sin\left(2\pi y\right)f(y)dy.$$  I used this to find that $\cos(2\pi x)+\sin (2\pi x)$ is an eigenvector.  I wasn't sure if this is the right track, and I am still confused about what to do from here.  (Specifically, once we find a bunch of $\lambda$ how do we prove that is all of them?)
 A: Let's start with where you're stuck ($\lambda$ is an eigenvalue):
$$\lambda f(x)=\cos(2\pi x)\int_0^1\cos\left(2\pi y\right)f(y)\mathrm dy+\sin\left(2\pi x\right)\int_0^1\sin\left(2\pi y\right)f(y)\mathrm dy$$
Let $c_1=\int_0^1\cos\left(2\pi y\right)f(y)\mathrm dy$ and $c_2=\int_0^1\sin\left(2\pi y\right)f(y)\mathrm dy$; $f(x)$ thus has the form
$$f(x)=\frac1{\lambda}\left(c_1\cos(2\pi x)+c_2\sin\left(2\pi x\right)\right)$$
We then assemble two equations: one where both sides of the equation are multiplied by $\cos(2\pi x)$, and one multiplied by $\sin(2\pi x)$, and then integrate both sides of the equation. I'll do it for $\cos(2\pi x)$:
$$\cos(2\pi x)f(x)=\frac1{\lambda}\left(c_1\cos(2\pi x)\cos(2\pi x)+c_2\sin\left(2\pi x\right)\cos(2\pi x)\right)$$
$$c_1=\frac1{\lambda}\left(c_1\int_0^1\cos(2\pi x)\cos(2\pi x)\mathrm dx+c_2\int_0^1\sin\left(2\pi x\right)\cos(2\pi x)\mathrm dx\right)$$
You should be able to recognize an (algebraic) eigenvalue equation at this point; the "secular equation" I was referring to in the comments is sometimes also referred to as the "characteristic polynomial".
