Can someone give me a clue about how to solve the b part ?
All I know is the self adjoint formula $$\langle ku,u\rangle = \lambda\langle u,u\rangle$$
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Do you know the definition of a self-adjoint? – Mhenni Benghorbal Feb 10 '16 at 21:37
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You can follow this technique. – Mhenni Benghorbal Feb 10 '16 at 21:52
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Since $K$ is self-adjoint, we have $\langle Ku,v\rangle=\langle u,Kv\rangle$. If $Ku=\lambda u$ and $u=v\ne0$, then $$ \lambda \langle u,u\rangle=\langle \lambda u,u\rangle=\langle u,\lambda u\rangle=\overline\lambda\langle u,u\rangle, $$ where the second identity comes from the fact that $K$ is self-adjoint. Since $\langle u,u\rangle=\|u\|^2\ne0$, we conclude that $\lambda =\overline\lambda$ and so $\lambda$ is real.
John B
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sorry i dint read from where second identity came, but now its clear. thank you. – Nithish Feb 10 '16 at 21:39