# Find the eigenvalues and eigenfunctions of integral operator

Find the eigenvalues and eigenfunctions of integral operator

$$Ku(x)= \int_0^\pi \sin(x)\sin(2y)u(y)\,\mathrm dy$$

I tried to use separable kernel to solve this, but I get a zero matrix $A$, please help, thanks!

Assuming that we are in the Hilbert space $H = L^2(0,\pi)$:
At first step we can rewrite the operator $K$ as $$Ku(x) = \sin(x) \int_{0}^{\pi} \sin(2y)u(y)\,\mathrm{d}y .$$ From that we can conclude that $\operatorname{ran} K = \operatorname{span}\{ \sin\}$ and that $$\ker K = \{\sin(2y)\}^{\perp} = \overline{\operatorname{span} \{\sin(nx):n\in\mathbb{N}\backslash\{2\}\} \cup \{\cos(nx):n\in\mathbb{N}\cup\{0\}\}}$$ Clearly every $u \in \ker K$ is a eigenfunction with eigenvalue $0$. For the remaining direction $u(x)=\sin(2x)$ we get $$Ku(x) = \sin(x) \int_0^\pi \sin^2(2y) \,\mathrm{d}y = \frac{\pi}{2} \sin(x)$$ So this is not an eigenfunction.
• Thanks a lot! So $\lambda=0$ is the only eigenvalue, and the eigenfunctions would be the functions orthogonal to the span, but why do you need to check sin(2x)? Aug 20, 2017 at 2:53
• You are right it is clear that only functions which are in the $\operatorname{ran} K$ can be eigenfunctions to an eigenvalue $\lambda \neq 0$. So this wasn't nessacery. Aug 20, 2017 at 8:49