# Prove that P is an orthogonal projection if and only if P is self adjoint.

1. Jul 24, 2009

### evilpostingmong

1. The problem statement, all variables and given/known data
Suppose P ∈ L(V) is such that P2 = P. Prove that P is an orthogonal
projection if and only if P is self-adjoint.

2. Relevant equations

3. The attempt at a solution
Let v be a vector in V. Let w be a vector in W and u be a vector in U and let U and W be subspaces of V where dim W+dim U=dimV. Let v=u+w. Now let P be an orthogonal projection onto u. We know that u and w are orthogonal so <u, w>=0. So P2=Pu=u. Since Pu=u, <Pu, w>=0. Now we know that Pw=0 since PP(v)=PP(u+w)=P(Pu+Pw)=P(u+0)=Pu therefore <u, Pw>=<u, 0>=0. Thus <Pu, w>=0=<u, Pw>.

Last edited: Jul 24, 2009
2. Jul 24, 2009

### Office_Shredder

Staff Emeritus
You have no reason to believe u and w are orthogonal. You need to be more clever picking your subspaces... in particular, try looking at U=P(V) the image of V, and W be the kernel of P. Then since P is an orthogonal projection you get what you need

3. Jul 24, 2009

### evilpostingmong

Let v be a vector in V. Let w be a vector in W and u be a vector in U and let U and W be subspaces of V where dim W+dim U=dimV. Let W=nullP. Let U=P(V). Now let P be an orthogonal projection onto u. So P2=Pu. Since Pu is orthogonal to w, <Pu, w>=0.
Since w is in nullP, Pw=0 so <u, Pw>=<u, 0>=0. So <Pu, w>=<u, Pw>.

4. Jul 24, 2009

### Office_Shredder

Staff Emeritus
You have to order your argument properly. START with: P is an orthogonal projection. THEN pick your U and W as U=P(V) W=Kernel(P).

Now you need to show for ANY v1 and v2 in V, <Pv1,v2> = <v1,Pv2>. You should be able to write the v's in terms of vectors in U and W.

After that you need to start working on going in the other direction (if P is self adjoint, then P is an orthogonal projection)

5. Jul 24, 2009

### evilpostingmong

Ok let P be an orthogonal projection. U and W are subspaces of V. U=P(V) and W=kernel(P).
Now let v1 and v2 be vectors in V. Now let v1=u1+w1 and let v2=u2+w2 where wi is in W and ui is in U. Now is this what you mean by in terms of vectors in U and W?

6. Jul 24, 2009

### Office_Shredder

Staff Emeritus
Yep. You should be able to prove that given the assumptions (P is an orthogonal projection) that you can write any vector uniquely like that

7. Jul 24, 2009

### evilpostingmong

Ok let P be an orthogonal projection. U and W are subspaces of V. U=P(V) and W=kernel(P).
Now let v1 and v2 be vectors in V. Now let v1=u1+w1 and let v2=u2+w2 where wi is in W and ui is in U. Now consider
<P(w1+u1), u2+w2> and <w1+ u1, P(u2+w2)>. Now factoring both seperately gives <Pw1+Pu1, u2+w2>
or <Pu1, u2+w2> or <Pu1, u2>+<Pu1, w2>. Doing the same for the other inner product gives <w1, Pu2>+<u1, Pu2>.
Now since wj is orthogonal to uj, <Puj, wj>=0. So we are left with <Pu1, u2> and <u1, Pu2>. Since
Puj=uj, we are left with <u1, u2> and <u1, u2>. Therefore <Pu1, u2>=<u1, Pu2>.

Last edited: Jul 24, 2009
8. Jul 25, 2009

### evilpostingmong

is it true that <Puj, wj>=0? I mean I always figured that kernelP contains
vectors perpendicular to those in U. And when you orthogonally project,
you take a line and if you can (exception is if you are dealing with
a single dimension) drop a "perpendicular" from that line and hit another one and
that "other one" is the projection. The "perpendicular" should be the
one in nullP. So by v=u+w, the w is the perpendicular, the u is the projection, and
the hypoteneuse is the sum of those two. If I'm wrong here, then I have a misconception

9. Jul 25, 2009

### Office_Shredder

Staff Emeritus
Since uj is in the image of P, you have that Puj = uj and hence <Puj, wj>=0

10. Jul 25, 2009

### evilpostingmong

now before I go on to prove the other direction, should I still assume that V=W+U?

11. Jul 25, 2009

### Office_Shredder

Staff Emeritus
Why don't you write down a couple different ways that you can prove something is an orthogonal projection and see which one looks easiest, that is what conditions are sufficient to be an orthogonal projection

12. Jul 26, 2009

### evilpostingmong

Oh hold on how silly of me. There has to be SOME null space of P as well as a range space.
After all, P is a projection, a type of transformation.

Ok so we have V=nullP+rangeP or V=W+U. <Pv1, v2>=<v1, Pv2> v1=/=v2.. Now vi=ui+wi. So <P(u1+w1), u2+w2>=<u1+w1, P(u2+w2)>. Or <P(u1)+P(w1), u2+w2>=<u1+w1, P(u2)+P(w2)>. Now since wj is in nullP, P(wj)=0 so we have <P(u1), u2+w2>=<u1+w1, P(u2)>. Factoring, we end up with <P(u1), u2>+<P(u1), w2>=<u1, P(u2)>+<w1, P(u2)>. Now <P(u1), w2>=/=<w1, P(u2)> unless <P(u1), w2> and <w1, P(u2)>=0. This can only be possible if P(uj) is orthogonal to wj so in this case
we end up with <P(u1), u2>=<u1, P(u2)> which we know to be true.