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Let $P$ be a linear operator on a Hilbert space $H$. If $\operatorname{range} P=(\ker P)^\perp$, is $P$ necessarily a projection, i.e., $P^2=P$?

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    Obviously no. If $P$ is invertible (but not the identity to avoid being a projection), you'll have this too.2012-09-25
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    I think my answer may give _all_ counterexamples. Did I miss any?2012-09-25

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