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Let $\phi : L_2 (\mathbb{R}) \to L_2 (\mathbb{R})$ be a real operator ,meaning all of it eigenvalues are real, and assume the set of eigenvectors form a basis for $h = L_2 (\mathbb{R})$.

Can $\phi$ has infinitely many linearly independent eigenvector belonging to the same eigenvalue ?

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Have you looked at the identity operator on $L^2(\Bbb{R})$ with eigenvalue 1?

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  • $\begingroup$ I looked many places except right under my nose :), thanks for pointing out. $\endgroup$ – onurcanbektas Oct 12 '17 at 11:42

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