AxiomOfChoice
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If A is a bounded operator on a Hilbert space H, isn't the following true of the residual spectrum \sigma_r(A):
\lambda \in \sigma_r(A) iff (\forall \psi \in H, \psi \neq 0)((\lambda - A) \psi \neq 0) iff \ker (\lambda - A) = \{0\} iff \lambda - A is injective?
So isn't the condition that \lambda isn't an eigenvalue equivalent to the condition \lambda - A is injective?
\lambda \in \sigma_r(A) iff (\forall \psi \in H, \psi \neq 0)((\lambda - A) \psi \neq 0) iff \ker (\lambda - A) = \{0\} iff \lambda - A is injective?
So isn't the condition that \lambda isn't an eigenvalue equivalent to the condition \lambda - A is injective?