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I was googling "Hilbert space" and was reading the associated Wikipedia page when I found this statement confusing :

"Let $V$ be a closed subspace of an Hilbert space $H$. Then the inclusion mapping $i_V : V\rightarrow H$ is the adjoint of the orthogonal projection $P_V : H\rightarrow V$".

I understand that means $\langle i_V(f),g\rangle=\langle f,P_V(g)\rangle $ for all $f\in V, \,g\in H$. But it is also known that an orthogonal projection is self-adjoint, so that we should have $P_V=P_V^\dagger=i_V$, which is not correct, but I can't explain why.

  • Could you explain where is my mistake ?

  • Moreover, do you know a proof for the mentioned adjoint property ?

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It would help if you cited your source more carefully. Note that (1) Your citation misses the word "adjoint" (2) the article suggested currently does not contain the word "injection". – Marc van Leeuwen Dec 22 '11 at 12:10
Oups, let me edit that. – Student Dec 22 '11 at 12:20
up vote 2 down vote accepted

Your confusion seems to be with the codomain of maps. The orthogonal projection $P_V$ cannot be self-adjoint because it is not a map $H\to H$ but rather $H\to V$. If you extend the codomain to $H$ then it is no longer the adjoint of the injection $i_V$ but rather of an extension of that map to all of $H$ (namely project onto $V$ first and then apply $i_V$), and this map happens to be identical to the "projection with codomain extended to $H$" itself.

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It's actually the confusion of Wikipedia, too (an outrageously badly written passage in which $P_V$ denotes a map $H \to H$ and $P_V : H \to V$ at the same time...) – t.b. Dec 22 '11 at 12:19
@t.b.: Right. But I've corrected it (somewhat) now. – Marc van Leeuwen Dec 22 '11 at 13:24
And could one of you give me an idea about how to prove the statement from Wikipedia ? – Student Dec 22 '11 at 17:30
@Student: Write (uniquely) $y=y_0+y_1$ with $y_0\in V$ and $y_1\in V^\perp$. Then $\langle i(x), y\rangle_H = \langle i(x), y_0+y_1\rangle_H = \langle x, \pi(y)\rangle_V+ \langle i(x),y_1\rangle_H = \langle x, \pi(y)\rangle_V$ since $i(x)\in V\perp y_1$. – Marc van Leeuwen Dec 22 '11 at 17:46

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