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

[evilpostingmong]

Homework Statement

Suppose P ∈ L(V) is such that P² = P. Prove that P is an orthogonal
projection if and only if P is self-adjoint.

Homework Equations

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>.

 

[Office_Shredder]

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

 

[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>.

 

[Office_Shredder]

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)

 

[evilpostingmong]

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)

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?

 

[Office_Shredder]

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

 

[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 separately 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>.

 

[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
about orthogonal projections.

 

[Office_Shredder]

Since u_j is in the image of P, you have that Pu_j = u_j and hence <Puj, wj>=0

 

[evilpostingmong]

Since u_j is in the image of P, you have that Pu_j = u_j and hence <Puj, wj>=0

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

 

[Office_Shredder]

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

 

[evilpostingmong]

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

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.