Results 1 to 5 of 5

Thread: 2 norm of an orthogonal projector

  1. #1
    Member
    Joined
    Feb 2009
    Posts
    98

    2 norm of an orthogonal projector

    Let P be a projector.
    Show that if $\displaystyle \left\|P \right\|_2 = 1, $ then P is an orthogonal projection.


    I know that for any projection P, $\displaystyle \left\|P \right\|_2 \geq 1, $ , and that if P is an orthogonal projection, then $\displaystyle \left\|P \right\|_2 = 1, $, but I am not sure how to prove the other way around.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    10
    Quote Originally Posted by math8 View Post
    Let P be a projector.
    Show that if $\displaystyle \left\|P \right\|_2 = 1, $ then P is an orthogonal projection.


    I know that for any projection P, $\displaystyle \left\|P \right\|_2 \geq 1, $ , and that if P is an orthogonal projection, then $\displaystyle \left\|P \right\|_2 = 1, $, but I am not sure how to prove the other way around.
    If P is not orthogonal then $\displaystyle (\ker P)^\perp$, the orthogonal complement of the kernel of P, is not equal to the range of P. Also, those two spaces have the same dimension. So there must exist a vector $\displaystyle x\in(\ker P)^\perp$ such that $\displaystyle Px\ne x$. Then $\displaystyle Px = (Px-x) + x$. But $\displaystyle Px-x\in\ker(P)$ so it is orthogonal to x. It follows (Pythagoras' theorem) that $\displaystyle \|Px\|^2 = \|Px-x\|^2 + \|x\|^2$. Since $\displaystyle Px-x\ne0$, that tells you that $\displaystyle \|Px\|>\|x\|$ and hence $\displaystyle \|P\|_2>1$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Feb 2009
    Posts
    98
    Quote Originally Posted by Opalg View Post
    Also, those two spaces have the same dimension. So there must exist a vector $\displaystyle x\in(\ker P)^\perp$ such that $\displaystyle Px\ne x$.
    I don't quite follow here, why is that true?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    10
    Quote Originally Posted by math8 View Post
    I don't quite follow here, why is that true?
    The subspaces $\displaystyle \text{ran}(P)$ and $\displaystyle (\ker(P))^\perp$ have the same dimension because they both have $\displaystyle \ker(P)$ as a complementary subspace. If $\displaystyle (\ker(P))^\perp \subseteq \text{ran}(P)$ then (because they have the same dimension) these two subspaces would have to coincide, which would imply that P is orthogonal. Since P is not orthogonal, there must be an element $\displaystyle x\in(\ker(P))^\perp$ such that $\displaystyle x\notin\text{ran}(P)$. Then $\displaystyle x\ne Px$.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Feb 2009
    Posts
    98
    Thanks a lot, now I got it
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: Feb 7th 2011, 11:08 PM
  2. Replies: 3
    Last Post: Jul 13th 2010, 06:37 PM
  3. 2 norm of a projector
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: May 7th 2010, 08:05 AM
  4. Replies: 2
    Last Post: Nov 7th 2009, 12:13 PM
  5. Projector and inverse of a vector.
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Oct 14th 2009, 03:02 PM

Search tags for this page

Search Tags


/mathhelpforum @mathhelpforum