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Orthogonal projection

  1. Mar 18, 2006 #1
    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"?
     
  2. jcsd
  3. Mar 18, 2006 #2

    matt grime

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    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
     
  4. Mar 18, 2006 #3
    So T(e)=e?

    What does it mean in terms of projections?
     
    Last edited: Mar 18, 2006
  5. Mar 19, 2006 #4

    matt grime

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    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
     
  6. Mar 19, 2006 #5
    What about the identity transformation? It's not an orthogonal projection of anything.
     
  7. Mar 20, 2006 #6

    matt grime

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    Yes it is, onto the subspace itself.
     
  8. Mar 20, 2006 #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?
     
    Last edited: Mar 20, 2006
  9. Mar 20, 2006 #8

    mathwonk

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    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.
     
  10. Mar 20, 2006 #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.
     
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