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$V$ is a finitte dimension vector space. If for some $P\in End(V)$ we have $P=P^2$ then $P$ is an orthogonal projection $\iff$ $P$ is self-adjoint.

I can show that since $P$ is idempotent then $V=ker(P)\oplus Im(P)$ and $P$ is the projection on its image parallel to its kernel. If $P$ is also self-adjoint I easily get that $ker(P)$ is orthogonal to $Im(P)$ and therefore $P$ is an orthogonal projection.

I'm not quite sure how to proceed in the other direction, thoughts?

Edit - Answer (Thanks to Verdruss)

As shown above $\forall v,v'\in V$ we have a unique decomposition $v=u+w,\ v'=u'+w'$ $\ \ u,u'\in U;\ \ w,w'\in W$ where $U$ and $W$ are orthogonal.

$\left\langle P(v)\mid v'\right\rangle =\left\langle P(u+w)\mid u'+w'\right\rangle =\left\langle u\mid u'+w'\right\rangle =\left\langle u\mid u'\right\rangle +\left\langle u\mid w'\right\rangle =\left\langle u\mid u'\right\rangle $

$\left\langle v\mid P(v')\right\rangle =\left\langle u+w\mid P(u'+w')\right\rangle =\left\langle u+w\mid u'\right\rangle =\left\langle u\mid u'\right\rangle +\left\langle w\mid u'\right\rangle =\left\langle u\mid u'\right\rangle $

And therefore we can conclude $P$ is self-adjoint

H.Rappeport
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  • $V$ is a hilbert space? – Marios Gretsas Jul 18 '17 at 12:08
  • Note that there are non-orthogonal projection as given in this wiki page. So the other direction shall be easier and you have done the forward direction wrongly: https://en.m.wikipedia.org/wiki/Projection_(linear_algebra)#Orthogonal_projections – Li Chun Min Jul 18 '17 at 12:14
  • @Ramiz Karaeski $V$ is a finite dimension vector space, I'll add that – H.Rappeport Jul 18 '17 at 12:14
  • I meant your definition of orthogonal direction is not accurate. You need more conditions. – Li Chun Min Jul 18 '17 at 12:18
  • @ Li Chun Min How so? My definition is that $P$ is an orthogonal projection on some subspace $W$ iff $P(v)=w\in W$ and also $v-w$ is orthogonal to $W$. If $P$ is the projection on $W$ parallel to $U$ and also $W$ and $U$ are orthogonal subspaces it seems obvious to me that $P$ is in fact an orthogonal projection – H.Rappeport Jul 18 '17 at 13:01
  • From your 1st paragraph, I thought your definition is that " ($P^2=P$ and $P$ is an endomorphism) iff $P$ is a orthogonal projection". – Li Chun Min Jul 18 '17 at 13:05
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    You have to show that $<P(u+w),u'+w'> = <u+w, P(u'+w')>$ holds, where $U+W=V$ is the orthogonal sum induced by your projection. This shouldn't be too hard. Or maybe you have some other definitions that I'm not aware of? – Verdruss Jul 18 '17 at 13:27

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