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I'm trying to prove that for given two dual maps $A : H \to H$ and $A^* : H^* \to H^*$ where $H = L_2(\mathbb{R})$, the set of all eigenvalues of $A$ is equal to the set of all eigenvalues of $A^*$.

Of course I can prove it by saying that after fixing a a basis in $H$ and the dual basis in $H^*$, the matrix of $A$ is the transpose of the matrix of $A^*$, and hence they have the same eigenvalues, but I'm particularly interested in a solution using different approach.

Similar to the answer given in this post by @Robert, I'm trying to prove result similarly for infinite dimensional vector space $L_2 (\mathbb{R})$.

However, in @Robert's answer, he assumes surjectivity which I cannot since I'm trying to prove for any linear operator $A$, so my question is how can we prove this result for also infinite dimensional vector space $H = L_2 (\mathbb{R})$ with a similar method ?

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This is not true. An operator and its adjoint can have different sets of eigenvalues. Since $L_2(\mathbb{R})$ is isomorphic to the sequence space $\ell_2$, it suffices to construct an example for $\ell_2$. Consider the forward shift operator $$Sx= (0,x_1,x_2,x_3,\dots)$$ whose adjoint is the backward shift $$S^*x= (x_2,x_3,x_4,x_5,\dots)$$ Since $S^*$ maps the sequence $(1,0,0,\dots)$ to zero, the number $0$ is an eigenvalue of $S^*$. But it is not an eigenvalue of $S$, as $S$ is injective.

Suggested reading: Spectrum (functional analysis)

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  • $\begingroup$ According to which inner product are they dual maps ? $\endgroup$ – onurcanbektas Oct 7 '17 at 6:23

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