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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!

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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.

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  • $\begingroup$ 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)? $\endgroup$
    – Areedd
    Aug 20, 2017 at 2:53
  • $\begingroup$ 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. $\endgroup$ Aug 20, 2017 at 8:49
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    $\begingroup$ you could accept the answer if it answers your question ;) $\endgroup$ Aug 20, 2017 at 20:37

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