| Thread Closed |
Orthogonal projection |
Share Thread | Thread Tools |
| Mar18-06, 03:09 PM | #1 |
|
|
Orthogonal projection
Let T in L(V) be an idempotent linear operator on a finite dimensional inner product space. What does it mean for T to be "the orthogonal projection onto its image"?
|
| Mar18-06, 03:45 PM | #2 |
|
Recognitions:
|
Every element in v is a combination e+f where e is in the image of T and f in the kernel and T(e+f)=e
|
| Mar18-06, 05:53 PM | #3 |
|
|
So T(e)=e?
What does it mean in terms of projections? |
| Mar19-06, 03:52 AM | #4 |
|
Recognitions:
|
Orthogonal projection
Eh? What do you think a projection onto a subspace is? To me it *is* an idempotent linear map. Then nice thing about having an inner product (non-degenerat) around is that there is an obvious choice of complementary subspace
|
| Mar19-06, 10:16 AM | #5 |
|
|
What about the identity transformation? It's not an orthogonal projection of anything.
|
| Mar20-06, 08:39 AM | #6 |
|
Recognitions:
|
Yes it is, onto the subspace itself.
|
| Mar20-06, 06:38 PM | #7 |
|
|
Just to be certain, is it equivalent to saying
[tex]T(v)=Proj_{im(T)}(v)[/tex] for all v? And not knowing what the image of T is a priori does not matter? And how does it follow that each vector in V is a combination of some vector in its image and its nullspace? |
| Mar20-06, 09:46 PM | #8 |
|
Recognitions:
|
a projection is amap onto a subspace, that sends every vector to a vector in the subspace, and leaves vectorsd that are already in the subspace where they are.
so if V is a vector space and X is a subspace we want to project on, and if Y is any complementary subspace, i.e. X and Y together generate V, and X and Y have only the zewro vector in common, then every vector in V can be written uniquely in the form x+y where x is in X and y is in Y. Then the map f sending x+y to x, is a projection onto X, "along" Y. Notice that f(f(v)) = v for all v, since once v gets into X it stays put. And also Y = ker(f), since vectors in Y go to zero. Indeed any linear map f such that f^2 = f is asuch a projection. f is called an "orthogonal" projection if Y = ker(f) is orthogonal to X = im(f).\ at least i think so, you should of course verify everything by proving all these either trivial or false statements. |
| Mar20-06, 09:48 PM | #9 |
|
|
I figured it out. Of course I assumed that what is meant by "orthogonal projection onto its image" is that [tex]T(v)=Proj_{im(T)}(v)[/tex]. Thanks for the help.
|
| Thread Closed |
| Thread Tools | |
Similar Threads for: Orthogonal projection
|
||||
| Thread | Forum | Replies | ||
| [SOLVED] Quick Question: Is this matrix an orthogonal projection? | Introductory Physics Homework | 2 | ||
| [SOLVED] Hilbert space & orthogonal projection | Calculus & Beyond Homework | 2 | ||
| Find the orthogonal projection | Calculus & Beyond Homework | 0 | ||
| Matlab: Orthogonal Projection | Math & Science Software | 0 | ||
| Orthogonal Projection Matrix | Calculus & Beyond Homework | 1 | ||