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Math Help - 2 norm of an orthogonal projector

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    2 norm of an orthogonal projector

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


    I know that for any projection P, \left\|P \right\|_2 \geq 1, , and that if P is an orthogonal projection, then \left\|P \right\|_2 = 1, , but I am not sure how to prove the other way around.
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    Quote Originally Posted by math8 View Post
    Let P be a projector.
    Show that if \left\|P \right\|_2 = 1, then P is an orthogonal projection.


    I know that for any projection P, \left\|P \right\|_2 \geq 1, , and that if P is an orthogonal projection, then \left\|P \right\|_2 = 1, , but I am not sure how to prove the other way around.
    If P is not orthogonal then (\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 x\in(\ker P)^\perp such that Px\ne x. Then Px = (Px-x) + x. But Px-x\in\ker(P) so it is orthogonal to x. It follows (Pythagoras' theorem) that \|Px\|^2 = \|Px-x\|^2 + \|x\|^2. Since Px-x\ne0, that tells you that \|Px\|>\|x\| and hence \|P\|_2>1.
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    Quote Originally Posted by Opalg View Post
    Also, those two spaces have the same dimension. So there must exist a vector x\in(\ker P)^\perp such that Px\ne x.
    I don't quite follow here, why is that true?
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    Quote Originally Posted by math8 View Post
    I don't quite follow here, why is that true?
    The subspaces \text{ran}(P) and (\ker(P))^\perp have the same dimension because they both have \ker(P) as a complementary subspace. If (\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 x\in(\ker(P))^\perp such that x\notin\text{ran}(P). Then x\ne Px.
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    Thanks a lot, now I got it
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