A projection operator on a Hilbert space $H$ is defined as operator that projects a vector $x$ of $H$ onto an closed subspace $S$ of $H$. Why the subspace $S$ has to be closed?
Projection operator and closed subspaces
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linear-algebra
hilbert-spaces
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0It depends: exactly w$h$at properties would you like your projection operator to have? – 2011-11-30
2 Answers
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$S = \{x: P(x) = x\}$, so if the projection $P$ is continuous $S$ must be closed.
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0We might project it onto $S$, but not with a continuous projection. – 2011-12-02
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Your definition of projection could be regarded redundant. However something interesting is given by the case when your subspace is not closed, but a non-closed dense subset. To which vector do project your $x \notin S$ then?
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0Can you give an example of such an $S$? – 2011-12-01