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Demonstrations on Projections

  1. Apr 21, 2010 #1
    Studying the introductory chapters of a Operator Theory book, I have found that the author seem to find a lot of demonstrations "easy" and not worthy of demonstrations. Yet, I don't have such ease as he has... For instance, one that has bugged me (on several books) is the proof that:
    where E is a projection and I is the identity matrix.
    It always follows stating that its is simple, then, to tell the set of all surjective projections of a linear transformation L.

    How are they related? I can't seem to find any way to prove/answer the above and its been a couple of days now.

    Can anyone shed some light in this? I'd really appreciate it!


  2. jcsd
  3. Apr 21, 2010 #2
    First, you may acquire some intuiton about it by drawing a picture of an orthogonal projection in [itex]\mathbb{R}^3[/itex].

    Second, consider the first equality range(I - E) = ker(E). Suppose y belongs to range(I - E): then you may write it, in terms of I - E, how? What must Ey be equal to? On the other hand, if Ey = 0, what is (E - I)y?
    Last edited: Apr 21, 2010
  4. Apr 21, 2010 #3

    Well... maybe the thing is that I don't fully understand what I-E is, as a matter of fact.
    One of the things I remember is that ker(E)={0}, the null vector. Or is it just for linear spaces?
  5. Apr 21, 2010 #4
    Consider a 2D plane P in [itex]\matbb{R}^3[/itex] and a vector v, not in P; then, if E is the orthogonal projection on P, then (I - E)v is the orthogonal projection on normal to the plane P.

    No, this is wrong. The kernel of a linear transformation [itex]T:V\rightarrow W[/itex] is the set:

    [tex]ker\left(T\right)=\left\{v\in V:Tv=0\right\}[/tex]

    Where the 0 is the null vector of W.

    Here's a hint:

    [tex]y \in R \left(I - E\right) \Leftrightarrow \exists x \in V:y=\left(I - E\right)x[/tex]

    Now remember that E is a projection, so E2=E.
  6. Apr 21, 2010 #5
    Oh yeah, I remember E2=E... used it plenty today!

    So, let me try to get things straight: [tex]R(I - E)[/tex] is the collectiong of all projections normal to the plane P (like the one in the example you gave)?
  7. Apr 21, 2010 #6
    No, R(I - E) is the range of the projection I - E (if E is a projection, I - E is also one). In euclidian spaces, you may interpret this as the subspace orthogonal to P, but the results you are trying to prove are more general than that; they are valid for linear spaces whithout a notion of orthogonality (usually given by an inner product).
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