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According to what I know, invariant subspace problem asks for if any operator of Banach space has a nontrivial closed invariant subspace. I am not sure what "nontrivial" means here. Of course $\{0\}$ and the whole space count as trivial subspace. What about kernel of the operator? If the operator has a nontrivial kernel(that is, not zero), does it count as nontrivial closed ,invariant subspace?

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Yes, $\ker(T)$ qualifies as a non-trivial closed invariant subspace whenever $T$ is not injective nor the zero map. Similarly, $\operatorname{im}(T)$ qualifies as a non-trivial closed invariant subspace whenever $T$ is not surjective and not-zero, so we may restrict our attention to bounded linear bijections from the Banach space to itself.

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