1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Orthogonal projectors

  1. Oct 1, 2009 #1
    Let P and Q be two m x m orthogonal projectors.
    We show a) ||P-Q||_2 <or eq. 1
    b)||P-Q||_2 < 1 implies the ranges of P and Q have equal dimensions.


    I think I must use the properties of orthogonal projectors. I guess Range(P) Inters Null(P) = {0} and Range(Q) Inters Null(Q) = {0}.
    Also ||P-Q||_2 = Max {|(P-Q)x|: ||x||=1|}. But I am not sure how to proceed from here.

    I am seeing somewhere this lemma: ||P-Q||_2 <1 iff Range(P) Inters Null(Q) = {0} and Range(Q) Inters Null(P) = {0}.

    Is it true? If yes why?
     
  2. jcsd
  3. Oct 1, 2009 #2
    I don't understand question (a).

    I also don't know the word 'inters' in this context. Do you mean 'contains'? If by Null(Q), you mean the set of vectors that it maps to zero, then it is most certainly not trivial for any non-trivial projection operator. In fact, one has [tex] V = Null(Q) \oplus Range(Q)[/tex].

    For example, in R3 for the standard basis, one has the projection operator

    [tex]P = \begin{bmatrix}1 & 0 & 0 \\0 & 1 & 0\\0 & 0 & 0 \end{bmatrix}[/tex]

    Null(P) is generated by (0, 0, 1)T and Range(P) is generated by (1, 0, 0)T and (0, 1, 0)T.
     
    Last edited: Oct 1, 2009
  4. Oct 2, 2009 #3
    Inters means Intersection here.
     
  5. Oct 2, 2009 #4
    That's funny. Inters means "to bury" in common usage. That's why I thought it meant contains.
     
  6. Oct 2, 2009 #5
    I figured out (b)

    [tex]v \in Ker(Q)\cap Im(P), |v| = 1 \ \Rightarrow \ (P - Q)v = v[/tex]

    One can move from there to

    [tex] ||P - Q||_{2} < 1 \ \Rightarrow Ker(Q)\cap Im(P) = \left\{0\right\}[/tex]

    Some standard projector properties and a little dimension counting close the deal.


    I still don't understand question (a) though....
     
  7. Oct 2, 2009 #6

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I read it as

    (a) Prove that [tex]||P - Q||_2 \leq 1[/tex]

    Not sure how to prove it.
     
  8. Oct 2, 2009 #7
    Heh, I'm new to ascii math. Should have gotten that....
     
  9. Oct 3, 2009 #8
    To handle the first one, factorize V as

    [tex]V = \left[ Im(P) \cap Im(Q) \right] \oplus \left[ Im(P) \cap Ker(Q) \right] \oplus
    \left[ Ker(P) \cap Im(Q) \right] \oplus \left[ Ker(P) \cap Ker(Q) \right] [/tex]

    So we can express a vector v = v1 + v2 + v3 +v4. We also have vivj = 0 if i isn't j by picking a basis.

    From here we compare ||(P - Q)v|| to ||v|| to get the more general conclusion that ||(P - Q)v|| <or eq ||v||




    I don't seem to be able to find a simpler way
     
    Last edited: Oct 4, 2009
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Orthogonal projectors
  1. Oblique projector (Replies: 1)

  2. Normal Projector (Replies: 5)

  3. Projector matrices (Replies: 8)

Loading...