Does an infinite dimensional operator $T$ having an eigenvalue of $0$ imply that $T^{-1}$ doesn't exist?

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I can't seem to find anywhere a discussion on whether or not the idea of a zero eigenvalue (value in the point spectrum) implies the inverse doesn't exists for an infinite dimensional operator.

We know for a finite operator, this is the case (a 0 eigenvalue means a matrix is not invertible). I am wondering if this same condition (or some other condition on the eigenvalues or spectrum) gives us that the inverse of an infinite dimensional operator doesn't exist.

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Yes. Having an eigenvalue of $0$ is equivalent to $\operatorname{ker} (T - 0I)$ is non-trivial, which is equivalent to $T$ not being injective. What breaks down is that $0$ not being an eigenvalue only implies $T$ is injective, not that $T^{-1}$ exists, as injectivity and bijectivity are no longer equivalent for linear operators on infinite-dimensional spaces.