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Let $A$ be an operator in Hilbert space, and let the elements in H be composed of an orthonormal system of eigenvectors $\phi_n$. Then any completely continuous operator satisfies the notion $$\lim_{n\longrightarrow \infty}\lambda_n=0$$.

And we have also that $$Ax=\sum \lambda_n c_n \phi_n$$

Would then this imply that, if H is infinite dimensional, then the image of $Ax$ is infinite too?

Thanks

Luthier415Hz
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    Take for example the identity operator $I$. – David Raveh Jan 17 '23 at 10:19
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    A completely continuous operator on Hilbert space may not have eigenvalues. Unless it is symmetric, or more generally normal. – Ryszard Szwarc Jan 17 '23 at 11:40
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    The range of $A$ can easily be finite dimensional if only finitely many eigenvalues are different from $0$. These finite rank operators are dense in the compact operators. – Jochen Jan 17 '23 at 17:21

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